Computer Laboratory Manual: Calculus for the Life Sciences - Rohan
Calculus for Life Sciences · 2013-01-22 · Calculus for Life Sciences MAT 1332 C Winter 2010 Jing...
Transcript of Calculus for Life Sciences · 2013-01-22 · Calculus for Life Sciences MAT 1332 C Winter 2010 Jing...
Calculus for Life SciencesMAT 1332 C Winter 2010
Jing Li
Department of Mathematics and StatisticsUniversity of Ottawa
March 8, 2010
Jing Li (UofO) MAT 1332 C March 8, 2010 1 / 39
Outline
1 Midterm 1 Grade Statistics
2 Feedback Questionaire
3 Linear Algebra I: Solving Linear Systems of Equations
4 Linear Algebra II: Vectors and MatricesDefinitionOperations
Basic Matrix OperationsMatrix-Vector MultiplicationMatrix-Matrix Multiplication
Jing Li (UofO) MAT 1332 C March 8, 2010 2 / 39
Midterm 1 Grade Statistics
Table: Midterm 1 Grade Statistics
Count: 169Average: 18.8Median: 19.5Maximum: 29.5Minimun: 2.0Standard Deviation: 5.91
Jing Li (UofO) MAT 1332 C March 8, 2010 3 / 39
Midterm 1 Grade Statistics
Midterm 1 Grade Statistics
Jing Li (UofO) MAT 1332 C March 8, 2010 4 / 39
Feedback Questionaire
Aspects that help learning
Algorithms
DGDs
Help available outside class
Examples in class
Assignments
Math help center
Review sessions
Jing Li (UofO) MAT 1332 C March 8, 2010 5 / 39
Feedback Questionaire
Suggestions:
Better English
Better classroom
Better blackboard presentation (notes )
Slow down (especially in the end of class)
Practice midterms
Better control of the class (too noisy)
Better organization and better introduction
More explanation
Application in real lives
DGDs should focus on similar problems for midterms and assignments.
get class more involved in the lecture.
More examples (difficult ones, close to tests and assignments, less parameters)
Better TA picking
do assignment questions in class
more office hours
Jing Li (UofO) MAT 1332 C March 8, 2010 6 / 39
Feedback Questionaire
Changes that can be implementedget help from me, TA, help center.NOTEs (online)Better lecture presentation (organization, introduction, examples,explanation, application)Slow down speedBetter control of the class (quite, get class more involved inlectures)Better questions in DGDs. (do hard assignment questions inDGDs)
Jing Li (UofO) MAT 1332 C March 8, 2010 7 / 39
Linear Algebra I: Solving Linear Systems of Equations
Reduced row-echelon form
Definition
leading entry, row-echelon form, and reduced row-echelon form
The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.
A matrix is in row-echelon form if the following three rules are true
(1) Rows of zeros are blew any nonzero row.(2) The leading entry of any row is to the right of any leading entry in
any row above it.(3) All entires in the column below a leading entry are zero.
A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition
(4) Each leading entry is 1.(5) All entries in the column above a leading entry are zero.
Jing Li (UofO) MAT 1332 C March 8, 2010 8 / 39
Linear Algebra I: Solving Linear Systems of Equations
Reduced row-echelon form
Definition
leading entry, row-echelon form, and reduced row-echelon form
The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.
A matrix is in row-echelon form if the following three rules are true
(1) Rows of zeros are blew any nonzero row.(2) The leading entry of any row is to the right of any leading entry in
any row above it.(3) All entires in the column below a leading entry are zero.
A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition
(4) Each leading entry is 1.(5) All entries in the column above a leading entry are zero.
Jing Li (UofO) MAT 1332 C March 8, 2010 8 / 39
Linear Algebra I: Solving Linear Systems of Equations
Reduced row-echelon form
Definition
leading entry, row-echelon form, and reduced row-echelon form
The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.
A matrix is in row-echelon form if the following three rules are true
(1) Rows of zeros are blew any nonzero row.
(2) The leading entry of any row is to the right of any leading entry inany row above it.
(3) All entires in the column below a leading entry are zero.
A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition
(4) Each leading entry is 1.(5) All entries in the column above a leading entry are zero.
Jing Li (UofO) MAT 1332 C March 8, 2010 8 / 39
Linear Algebra I: Solving Linear Systems of Equations
Reduced row-echelon form
Definition
leading entry, row-echelon form, and reduced row-echelon form
The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.
A matrix is in row-echelon form if the following three rules are true
(1) Rows of zeros are blew any nonzero row.(2) The leading entry of any row is to the right of any leading entry in
any row above it.
(3) All entires in the column below a leading entry are zero.
A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition
(4) Each leading entry is 1.(5) All entries in the column above a leading entry are zero.
Jing Li (UofO) MAT 1332 C March 8, 2010 8 / 39
Linear Algebra I: Solving Linear Systems of Equations
Reduced row-echelon form
Definition
leading entry, row-echelon form, and reduced row-echelon form
The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.
A matrix is in row-echelon form if the following three rules are true
(1) Rows of zeros are blew any nonzero row.(2) The leading entry of any row is to the right of any leading entry in
any row above it.(3) All entires in the column below a leading entry are zero.
A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition
(4) Each leading entry is 1.(5) All entries in the column above a leading entry are zero.
Jing Li (UofO) MAT 1332 C March 8, 2010 8 / 39
Linear Algebra I: Solving Linear Systems of Equations
Reduced row-echelon form
Definition
leading entry, row-echelon form, and reduced row-echelon form
The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.
A matrix is in row-echelon form if the following three rules are true
(1) Rows of zeros are blew any nonzero row.(2) The leading entry of any row is to the right of any leading entry in
any row above it.(3) All entires in the column below a leading entry are zero.
A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition
(4) Each leading entry is 1.(5) All entries in the column above a leading entry are zero.
Jing Li (UofO) MAT 1332 C March 8, 2010 8 / 39
Linear Algebra I: Solving Linear Systems of Equations
Reduced row-echelon form
Definition
leading entry, row-echelon form, and reduced row-echelon form
The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.
A matrix is in row-echelon form if the following three rules are true
(1) Rows of zeros are blew any nonzero row.(2) The leading entry of any row is to the right of any leading entry in
any row above it.(3) All entires in the column below a leading entry are zero.
A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition
(4) Each leading entry is 1.
(5) All entries in the column above a leading entry are zero.
Jing Li (UofO) MAT 1332 C March 8, 2010 8 / 39
Linear Algebra I: Solving Linear Systems of Equations
Reduced row-echelon form
Definition
leading entry, row-echelon form, and reduced row-echelon form
The leading entry of a row in a matrix is the leftmost nonzero coefficient in thatrow.
A matrix is in row-echelon form if the following three rules are true
(1) Rows of zeros are blew any nonzero row.(2) The leading entry of any row is to the right of any leading entry in
any row above it.(3) All entires in the column below a leading entry are zero.
A matrix is in reduced row-echelon form if it is in row-echelon form and inaddition
(4) Each leading entry is 1.(5) All entries in the column above a leading entry are zero.
Jing Li (UofO) MAT 1332 C March 8, 2010 8 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 5.
Categorize the following systems according to the above definition:241 2 00 3 10 0 2
35 ,
242 0 2 40 1 1 00 0 0 7
35 ,
241 0 20 1 10 0 0
35 ,
241 0 2 00 1 −3 00 0 0 1
35 ,
242 0 21 1 10 3 0
35Solution:
row-echelon form but not reduced: (1), (2)
reduced row-echelon form: (3), (4)
neither row-echelon nor reduced row-echelon form: (5)
Jing Li (UofO) MAT 1332 C March 8, 2010 9 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 5.
Categorize the following systems according to the above definition:241 2 00 3 10 0 2
35 ,
242 0 2 40 1 1 00 0 0 7
35 ,
241 0 20 1 10 0 0
35 ,
241 0 2 00 1 −3 00 0 0 1
35 ,
242 0 21 1 10 3 0
35Solution:
row-echelon form but not reduced: (1), (2)
reduced row-echelon form: (3), (4)
neither row-echelon nor reduced row-echelon form: (5)
Jing Li (UofO) MAT 1332 C March 8, 2010 9 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 5.
Categorize the following systems according to the above definition:241 2 00 3 10 0 2
35 ,
242 0 2 40 1 1 00 0 0 7
35 ,
241 0 20 1 10 0 0
35 ,
241 0 2 00 1 −3 00 0 0 1
35 ,
242 0 21 1 10 3 0
35Solution:
row-echelon form but not reduced: (1), (2)
reduced row-echelon form: (3), (4)
neither row-echelon nor reduced row-echelon form: (5)
Jing Li (UofO) MAT 1332 C March 8, 2010 9 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35
−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 5
0 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50
− 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2
− 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 |
− 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35
R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→
24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −10
0 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100
0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0
5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 |
20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35
(−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 5
0 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50
2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2
3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 |
100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 10
0 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100
0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0
1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 |
4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35
−R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→
24 1 1 0 | 10 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24
1 1 0 | 10 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1
1 0 | 10 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1
0 | 10 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 |
10 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0
2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2
0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 |
− 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35
12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 1
0 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10
1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1
0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 |
− 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35
−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24
1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1
0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0
0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 |
20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35
Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 6: Exactly One Solution
8<:x1 + x2 + x3 = 52x1 − x3 = 0
x2 + 4x3 = 15
Solution: Using the matrix notation,24 1 1 1 | 52 0 −1 | 00 1 4 | 15
35−2R1 + R2−−−−−−−→
24 1 1 1 | 50 − 2 − 3 | − 100 1 4 | 15
35R2 + 2R3−−−−−−→24 1 1 1 | 50 −2 −3 | −100 0 5 | 20
35 (−1)R2;15 R3
−−−−−−−→
24 1 1 1 | 50 2 3 | 100 0 1 | 4
35 −R3 + R1;−3R3 + R2−−−−−−−−−→24 1 1 0 | 1
0 2 0 | − 20 0 1 | 4
35 12
R2
−−→
24 1 1 0 | 10 1 0 | − 10 0 1 | 4
35−R2 + R1−−−−−−→
24 1 0 0 | 20 1 0 | −10 0 1 | 4
35Hence, we have solution x1 = 2, x2 = −1, x3 = 4 or (2,−1, 4).
Jing Li (UofO) MAT 1332 C March 8, 2010 10 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35
−2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 3
0 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30
− 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2
1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 |
− 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 4
0 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40
− 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2
1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 |
− 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35
−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→
24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −4
0 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40
0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0
0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 |
2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35
The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 7: No Solution
8<:2x1 − x2 + x3 = 34x1 − 4x2 + 3x3 = 22x1 − 3x2 + 2x3 = 1
Solution:24 2 −1 1 | 34 −4 3 | 22 −3 2 | 1
35 −2R1 + R2
−R1 + R3−−−−−−−−−→
24 2 −1 1 | 30 − 2 1 | − 40 − 2 1 | − 2
35−R2 + R3−−−−−−→24 2 −1 1 | 30 −2 1 | −40 0 0 | 2
35The last equation: 0 · x1 + 0 · x2 + 0 · x3 = 2, is wrong statement.
This is clearly impossible.
This means that this system does not have a solution.
Hence, this system is inconsistent.
Jing Li (UofO) MAT 1332 C March 8, 2010 11 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35
−R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 4
0 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40
1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1
2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 |
20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 2
0 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20
0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0
0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 |
0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35
The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35
The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement,
and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t ,
then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 8: Infinitely Many Solutions
8<:x1 − 3x2 + x3 = 4x1 − 2x2 + 3x3 = 62x1 − 6x2 + 2x3 = 8
Solution:24 1 −3 1 | 41 −2 3 | 62 −6 2 | 8
35 −R1 + R2
−2R1 + R3−−−−−−−−−→
24 1 −3 1 | 40 1 2 | 20 0 0 | 0
35The last row reads: 0 · x1 + 0 · x2 + 0 · x3 = 0.
This equation is a correct statement, and is true for all values of x1, x2, x3.
The second equation is x2 + 2x3 = 2.
We introduce a free variable by denoting x3 = t , then x2 = 2− 2x3 = 2− 2t .
The first equation: x1 − 3x2 + x3 = 4, =⇒x1 = 4 + 3x2 − x3 = 4 + 3(2− 2t)− t = 10− 7t .
Hence, the solution set is {(x1, x2, x3) : x1 = 10− 7t , x2 = 2− 2t , x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 12 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–
−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 1
0 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10
− 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3
2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 |
1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–
This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.
Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.
Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.
We set x3 = t as a free variable, then we get x2 = 23 x3 − 1
3 = 23 t − 1
3 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 9: Underdetermined System: A System has fewer equations than variables
2x1 + 2x2 − x3 = 12x1 − x2 + x3 = 2
Solution: »2 2 −1 | 12 −1 1 | 2
–−R1 + R2−−−−−−→
»2 2 −1 | 10 − 3 2 | 1
–This time, the last row reads −3x2 + 2x3 = 1.Whatever we choose for x3 we can always find a x2 to make the equation true.Hence this system has infinitely many solutions.We set x3 = t as a free variable, then we get x2 = 2
3 x3 − 13 = 2
3 t − 13 .
We plug these back to the first equation to solve for x1:
x1 = −x2 +12
x3 +12
= −(23
t − 13
) +12
t +12
= − t6
+56
.
Hence, the solution set is
{(x1, x2, x3) : x1 = − t6
+56
, x2 =23
t − 13
, x3 = t , t ∈ R}
Jing Li (UofO) MAT 1332 C March 8, 2010 13 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35
−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 2
0 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20
− 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 |
1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35
−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 2
0 1 | − 12
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20
1 | − 12
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 |
− 12
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35
R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24
2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2
0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 |
12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1
0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 |
52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35
12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24
1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1
0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 |
14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35
x1 = 14 from the first row, x1 = 5
2 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 10: Overdetermined System: a system has more equations than variables
8<:2x1 − x2 = 1x1 + x2 = 2x1 − x2 = 3
Solution:24 2 −1 | 11 1 | 21 −1 | 3
35−R2 + R3−−−−−−→
24 2 −1 | 11 1 | 20 − 2 | 1
35−12
R3
−−−−→
24 2 −1 | 11 1 | 20 1 | − 1
2
35R3 + R1
−R3 + R2−−−−−−−−→
24 2 0 | 12
1 0 | 52
0 1 | − 12
35 12
R1
−−→
24 1 0 | 14
1 0 | 52
0 1 | − 12
35x1 = 1
4 from the first row, x1 = 52 from the second row.
Since 14 6=
52 , there cannot be a solution.
The system is inconsistent and has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 14 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–
2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −3
0 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30
2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 |
0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .
If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.
if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.
There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra I: Solving Linear Systems of Equations
Example 11: Problem with parameters
x1 + hx2 = −3−2x1 + 4x2 = 6
have (1) a unique solution, (2) infinitely many solution, and (3) no solution?Solution: »
1 h | −3−2 4 | 6
–2R1 + R2−−−−−−→
»1 h | −30 2h + 4 | 0
–
The last row gives the equation: (2h + 4)x2 = 0.
If 2h + 4 = 0 (i.e., h = −2), then 0 · x2 = 0, x2 is a free variable.Then x2 = t , and x1 = −3− hx2 = −3− 2t .If 2h + 4 6= 0 (i.e., h 6= −2 ), then the only way to satisfy thisequation is x2 = 0, then x1 = −3.
Hence,
if h 6= −2, then there is a unique solution.if h = −2, then there are infinitely many solutions.There is so such of value of h so that the system has no solution.
Jing Li (UofO) MAT 1332 C March 8, 2010 15 / 39
Linear Algebra II: Vectors and Matrices Definition
Vectors and Matrices
Definition
A matrix is a rectangular array of numbers A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 .
The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.
Note:
If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].
If m = n, then A is a square matrix. eg. a 3× 3 square matrix:
24−1 3 00 1 −15 4 3
35If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.
For the above example, the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.
Jing Li (UofO) MAT 1332 C March 8, 2010 16 / 39
Linear Algebra II: Vectors and Matrices Definition
Vectors and Matrices
Definition
A matrix is a rectangular array of numbers A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 .
The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.
Note:
If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].
If m = n, then A is a square matrix. eg. a 3× 3 square matrix:
24−1 3 00 1 −15 4 3
35If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.
For the above example, the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.
Jing Li (UofO) MAT 1332 C March 8, 2010 16 / 39
Linear Algebra II: Vectors and Matrices Definition
Vectors and Matrices
Definition
A matrix is a rectangular array of numbers A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 .
The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.
Note:
If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].
If m = n, then A is a square matrix. eg. a 3× 3 square matrix:
24−1 3 00 1 −15 4 3
35If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.
For the above example, the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.
Jing Li (UofO) MAT 1332 C March 8, 2010 16 / 39
Linear Algebra II: Vectors and Matrices Definition
Vectors and Matrices
Definition
A matrix is a rectangular array of numbers A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 .
The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.
Note:
If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].
If m = n, then A is a square matrix. eg. a 3× 3 square matrix:
24−1 3 00 1 −15 4 3
35
If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.
For the above example, the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.
Jing Li (UofO) MAT 1332 C March 8, 2010 16 / 39
Linear Algebra II: Vectors and Matrices Definition
Vectors and Matrices
Definition
A matrix is a rectangular array of numbers A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 .
The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.
Note:
If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].
If m = n, then A is a square matrix. eg. a 3× 3 square matrix:
24−1 3 00 1 −15 4 3
35If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.
For the above example, the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.
Jing Li (UofO) MAT 1332 C March 8, 2010 16 / 39
Linear Algebra II: Vectors and Matrices Definition
Vectors and Matrices
Definition
A matrix is a rectangular array of numbers A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 .
The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.
Note:
If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].
If m = n, then A is a square matrix. eg. a 3× 3 square matrix:
24−1 3 00 1 −15 4 3
35If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.
For the above example,
the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.
Jing Li (UofO) MAT 1332 C March 8, 2010 16 / 39
Linear Algebra II: Vectors and Matrices Definition
Vectors and Matrices
Definition
A matrix is a rectangular array of numbers A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 .
The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.
Note:
If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].
If m = n, then A is a square matrix. eg. a 3× 3 square matrix:
24−1 3 00 1 −15 4 3
35If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.
For the above example, the diagonal elements are −1, 1, 3,
and the trace istr(A) = −1 + 1 + 3 = 3.
Jing Li (UofO) MAT 1332 C March 8, 2010 16 / 39
Linear Algebra II: Vectors and Matrices Definition
Vectors and Matrices
Definition
A matrix is a rectangular array of numbers A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 .
The elements aij of the matrix A are called entries. If the matrix has m rows and ncolumns, it is called an m × n matrix.
Note:
If the size of the matrix A is clear, we can use the shorthand notation A = [aij ].
If m = n, then A is a square matrix. eg. a 3× 3 square matrix:
24−1 3 00 1 −15 4 3
35If A is a square matrix, then the elements aii are called the diagonal elementsand their sum are called the trace of A.
For the above example, the diagonal elements are −1, 1, 3, and the trace istr(A) = −1 + 1 + 3 = 3.
Jing Li (UofO) MAT 1332 C March 8, 2010 16 / 39
Linear Algebra II: Vectors and Matrices Definition
A 1× n matrix is called a row vector: [c1, c2, · · · , cn]. eg. a 1× 4 row vector:[1 3 0 5].
An m × 1 matrix is called a column vector:
26664b1
b2...
bm
37775. eg. a 3× 1 column vector:
24274
35.
Two matrices A = [aij ], B = [bij ] are said to be equal if they have the samedimension and if for all i, j , we have aij = bij .
the zero matrix:
2640 · · · 0...
. . ....
0 · · · 0
375, the size of this matrix is usually clear from the
context.
the identity matrix:
266641 0 · · · 00 1 · · · 0...
.... . .
...0 0 · · · 1
37775, the size of this matrix is usually clear
from the context.
Jing Li (UofO) MAT 1332 C March 8, 2010 17 / 39
Linear Algebra II: Vectors and Matrices Definition
A 1× n matrix is called a row vector: [c1, c2, · · · , cn]. eg. a 1× 4 row vector:[1 3 0 5].
An m × 1 matrix is called a column vector:
26664b1
b2...
bm
37775. eg. a 3× 1 column vector:
24274
35.
Two matrices A = [aij ], B = [bij ] are said to be equal if they have the samedimension and if for all i, j , we have aij = bij .
the zero matrix:
2640 · · · 0...
. . ....
0 · · · 0
375, the size of this matrix is usually clear from the
context.
the identity matrix:
266641 0 · · · 00 1 · · · 0...
.... . .
...0 0 · · · 1
37775, the size of this matrix is usually clear
from the context.
Jing Li (UofO) MAT 1332 C March 8, 2010 17 / 39
Linear Algebra II: Vectors and Matrices Definition
A 1× n matrix is called a row vector: [c1, c2, · · · , cn]. eg. a 1× 4 row vector:[1 3 0 5].
An m × 1 matrix is called a column vector:
26664b1
b2...
bm
37775. eg. a 3× 1 column vector:
24274
35.
Two matrices A = [aij ], B = [bij ] are said to be equal if they have the samedimension and if for all i, j , we have aij = bij .
the zero matrix:
2640 · · · 0...
. . ....
0 · · · 0
375, the size of this matrix is usually clear from the
context.
the identity matrix:
266641 0 · · · 00 1 · · · 0...
.... . .
...0 0 · · · 1
37775, the size of this matrix is usually clear
from the context.
Jing Li (UofO) MAT 1332 C March 8, 2010 17 / 39
Linear Algebra II: Vectors and Matrices Definition
A 1× n matrix is called a row vector: [c1, c2, · · · , cn]. eg. a 1× 4 row vector:[1 3 0 5].
An m × 1 matrix is called a column vector:
26664b1
b2...
bm
37775. eg. a 3× 1 column vector:
24274
35.
Two matrices A = [aij ], B = [bij ] are said to be equal if they have the samedimension and if for all i, j , we have aij = bij .
the zero matrix:
2640 · · · 0...
. . ....
0 · · · 0
375, the size of this matrix is usually clear from the
context.
the identity matrix:
266641 0 · · · 00 1 · · · 0...
.... . .
...0 0 · · · 1
37775, the size of this matrix is usually clear
from the context.
Jing Li (UofO) MAT 1332 C March 8, 2010 17 / 39
Linear Algebra II: Vectors and Matrices Definition
A 1× n matrix is called a row vector: [c1, c2, · · · , cn]. eg. a 1× 4 row vector:[1 3 0 5].
An m × 1 matrix is called a column vector:
26664b1
b2...
bm
37775. eg. a 3× 1 column vector:
24274
35.
Two matrices A = [aij ], B = [bij ] are said to be equal if they have the samedimension and if for all i, j , we have aij = bij .
the zero matrix:
2640 · · · 0...
. . ....
0 · · · 0
375, the size of this matrix is usually clear from the
context.
the identity matrix:
266641 0 · · · 00 1 · · · 0...
.... . .
...0 0 · · · 1
37775, the size of this matrix is usually clear
from the context.
Jing Li (UofO) MAT 1332 C March 8, 2010 17 / 39
Linear Algebra II: Vectors and Matrices Definition
A 1× n matrix is called a row vector: [c1, c2, · · · , cn]. eg. a 1× 4 row vector:[1 3 0 5].
An m × 1 matrix is called a column vector:
26664b1
b2...
bm
37775. eg. a 3× 1 column vector:
24274
35.
Two matrices A = [aij ], B = [bij ] are said to be equal if they have the samedimension and if for all i, j , we have aij = bij .
the zero matrix:
2640 · · · 0...
. . ....
0 · · · 0
375, the size of this matrix is usually clear from the
context.
the identity matrix:
266641 0 · · · 00 1 · · · 0...
.... . .
...0 0 · · · 1
37775, the size of this matrix is usually clear
from the context.Jing Li (UofO) MAT 1332 C March 8, 2010 17 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix Addition and Scalar Multiplication
Definition
Matrix AdditionSuppose that A = [aij ] and B = [bij ] are two m × n matrices. Then
C = A + B
is an m × n matrix with entries
cij = aij + bij , for 1 ≤ i ≤ m, 1 ≤ j ≤ n.
Definition
Scalar MultiplicationSuppose that A = [aij ] is an m × n matrix and k is a number. Then
kA = [kaij ]
for 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Jing Li (UofO) MAT 1332 C March 8, 2010 18 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix Addition and Scalar Multiplication
Definition
Matrix AdditionSuppose that A = [aij ] and B = [bij ] are two m × n matrices. Then
C = A + B
is an m × n matrix with entries
cij = aij + bij , for 1 ≤ i ≤ m, 1 ≤ j ≤ n.
Definition
Scalar MultiplicationSuppose that A = [aij ] is an m × n matrix and k is a number. Then
kA = [kaij ]
for 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Jing Li (UofO) MAT 1332 C March 8, 2010 18 / 39
Linear Algebra II: Vectors and Matrices Operations
Note:
What are the sizes of C = A + B and kA?
C is an m × n matrix.kA is the same size as A. (m × n matrix)
In the field of Linear Algebra, a number is often called a scalar.
Jing Li (UofO) MAT 1332 C March 8, 2010 19 / 39
Linear Algebra II: Vectors and Matrices Operations
Note:
What are the sizes of C = A + B and kA?
C is an m × n matrix.kA is the same size as A. (m × n matrix)
In the field of Linear Algebra, a number is often called a scalar.
Jing Li (UofO) MAT 1332 C March 8, 2010 19 / 39
Linear Algebra II: Vectors and Matrices Operations
Note:
What are the sizes of C = A + B and kA?
C is an m × n matrix.kA is the same size as A. (m × n matrix)
In the field of Linear Algebra, a number is often called a scalar.
Jing Li (UofO) MAT 1332 C March 8, 2010 19 / 39
Linear Algebra II: Vectors and Matrices Operations
Note:
What are the sizes of C = A + B and kA?
C is an m × n matrix.kA is the same size as A. (m × n matrix)
In the field of Linear Algebra, a number is often called a scalar.
Jing Li (UofO) MAT 1332 C March 8, 2010 19 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.
Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»
8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8
10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10
123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 12
3 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123
3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3
3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–
2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»
− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6
− 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6
− 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 6
5 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65
7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7
9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–
3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»
5 10 1520 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5
10 1520 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10
1520 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20
25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25
30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–
4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»
3 6 912 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3
6 912 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6
912 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12
15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15
18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»
14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14
16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16
18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18
− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2
− 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4
− 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»
17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17
22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22
2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 27
10 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710
11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11
12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 1: Consider the following two 2× 3-matrices,
A =
»1 2 34 5 6
–, B =
»7 8 9−1 −2 −3
–,
Find A + B, A− B, 5A, 3A + 2B.Solution:
1 A + B =
»1 2 34 5 6
–+
»7 8 9−1 −2 −3
–=
»8 10 123 3 3
–2 A− B =
»1 2 34 5 6
–−»
7 8 9−1 −2 −3
–=
»− 6 − 6 − 65 7 9
–3 5A =
»5 10 15
20 25 30
–4 3A + 2B =
»3 6 9
12 15 18
–+
»14 16 18− 2 − 4 − 6
–=
»17 22 2710 11 12
–
Jing Li (UofO) MAT 1332 C March 8, 2010 20 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:
What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? –
AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.
What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? –
a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.
What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? –
a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.
Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric.
e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
The Transpose of a Matrix
Definition
Suppose that A = [aij ] is an m × n matrix. The transpose of A, denoted by AT , is ann ×m matrix with entries
AT = [aji ]
which is obtained from A by interchanging rows and columns, or, loosely speaking, byflipping the matrix along this diagonal.
Note:What is the size of AT ? – AT is n ×m if A is m × n.What is the transpose of a column vector? – a row vector.What is the transpose of a row vector? – a column vector.Like wise, the first column of AT is the first row of A; the second column of AT isthe second row of A, etc..
If AT = A, then A is symmetric. e.g. A =
24 1 2 32 4 63 6 5
35, then AT = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 21 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 2: Consider
A =
»1 2 34 5 6
–, B =
ˆ1 2
˜, C =
»45
–,
Find AT , BT , CT .Solution:
AT =
24 1 42 53 6
35BT =
»12
–CT =
ˆ4 5
˜
Jing Li (UofO) MAT 1332 C March 8, 2010 22 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 2: Consider
A =
»1 2 34 5 6
–, B =
ˆ1 2
˜, C =
»45
–,
Find AT , BT , CT .Solution:
AT =
24 1 42 53 6
35
BT =
»12
–CT =
ˆ4 5
˜
Jing Li (UofO) MAT 1332 C March 8, 2010 22 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 2: Consider
A =
»1 2 34 5 6
–, B =
ˆ1 2
˜, C =
»45
–,
Find AT , BT , CT .Solution:
AT =
24 1 42 53 6
35BT =
»12
–
CT =ˆ
4 5˜
Jing Li (UofO) MAT 1332 C March 8, 2010 22 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 2: Consider
A =
»1 2 34 5 6
–, B =
ˆ1 2
˜, C =
»45
–,
Find AT , BT , CT .Solution:
AT =
24 1 42 53 6
35BT =
»12
–CT =
ˆ4 5
˜
Jing Li (UofO) MAT 1332 C March 8, 2010 22 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix-Vector Multiplication
Definition
If the matrix A has n columns and the column vector x has n rows, then the productAx is defined as follows:
Ax =
26664a11x1 + a12x2 + · · ·+ a1nxn
a21x1 + a22x2 + · · ·+ a2nxn...
am1x1 + am2x2 + · · ·+ amnxn
37775where,
A =
26664a11 a12 · · · a1n
a21 a22 · · · a2n...
.... . .
...am1 am2 · · · amn
37775 , x =
26664x1
x2...
xn
37775 .
Jing Li (UofO) MAT 1332 C March 8, 2010 23 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix-Vector Multiplication
Note:
How to remember this definition?
simply think about linear systems of equations: Ax = b,a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2
...am1x1 + am2x2 + · · ·+ amnxn = bm
A: coefficient matrix of the linear system.
x =
x1x2...
xn
, unknown variable column vector. b =
b1b2...
bm
.
What is the size of Ax? –m × 1 column vector.
Jing Li (UofO) MAT 1332 C March 8, 2010 24 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix-Vector Multiplication
Note:
How to remember this definition?
simply think about linear systems of equations: Ax = b,a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2
...am1x1 + am2x2 + · · ·+ amnxn = bm
A: coefficient matrix of the linear system.
x =
x1x2...
xn
, unknown variable column vector. b =
b1b2...
bm
.
What is the size of Ax? –m × 1 column vector.
Jing Li (UofO) MAT 1332 C March 8, 2010 24 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix-Vector Multiplication
Note:
How to remember this definition?
simply think about linear systems of equations: Ax = b,a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2
...am1x1 + am2x2 + · · ·+ amnxn = bm
A: coefficient matrix of the linear system.
x =
x1x2...
xn
, unknown variable column vector. b =
b1b2...
bm
.
What is the size of Ax? –m × 1 column vector.
Jing Li (UofO) MAT 1332 C March 8, 2010 24 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix-Vector Multiplication
Note:
How to remember this definition?
simply think about linear systems of equations: Ax = b,a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2
...am1x1 + am2x2 + · · ·+ amnxn = bm
A: coefficient matrix of the linear system.
x =
x1x2...
xn
, unknown variable column vector. b =
b1b2...
bm
.
What is the size of Ax? –
m × 1 column vector.
Jing Li (UofO) MAT 1332 C March 8, 2010 24 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix-Vector Multiplication
Note:
How to remember this definition?
simply think about linear systems of equations: Ax = b,a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2
...am1x1 + am2x2 + · · ·+ amnxn = bm
A: coefficient matrix of the linear system.
x =
x1x2...
xn
, unknown variable column vector. b =
b1b2...
bm
.
What is the size of Ax? –m × 1 column vector.
Jing Li (UofO) MAT 1332 C March 8, 2010 24 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»
1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1
+ 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2
+ 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 3
4 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1
+ 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2
+ 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»
1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»14
32
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775
=
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»
1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)
4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–
=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»
724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»7
24
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35
=ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
=
6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–
=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»
1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 2
4 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–
=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»
36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»3
6
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–
=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24
1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)
4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)
3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35
=
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24
− 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3
− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7
− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 3: Calculate
1
»1 2 34 5 6
–24 123
35 =
»1 · 1 + 2 · 2 + 3 · 34 · 1 + 5 · 2 + 6 · 3
–=
»1432
–
2
»1 2 3 74 5 6 8
–2664123−1
3775 =
»1 · 1 + 2 · 2 + 3 · 3 + 7 · (−1)4 · 1 + 5 · 2 + 6 · 3 + 8 · (−1)
–=
»724
–
3ˆ
1 2 3˜ 24 1−23
35 =ˆ
1 · 1 + 2 · (−2) + 3 · 3˜
= 6
4
»1 24 5
– »−12
–=
»1 · (−1) + 2 · 24 · (−1) + 5 · 2
–=
»36
–
5
24 1 24 53 6
35» 1−2
–=
24 1 · 1 + 2 · (−2)4 · 1 + 5 · (−2)3 · 1 + 6 · (−2)
35 =
24 − 3− 7− 9
35
Jing Li (UofO) MAT 1332 C March 8, 2010 25 / 39
Linear Algebra II: Vectors and Matrices Operations
Matrix-Matrix Multiplication
Definition
Suppose that A = [aij ] is an m × n matrix and B = [bij ] is an n × k matrix. Then
C = AB
is an m × k matrix with
cij = ai1b1j + ai2b2j + · · ·+ ainbnj =nX
l=1
ailblj
for 1 ≤ i ≤ m and 1 ≤ j ≤ k.
Jing Li (UofO) MAT 1332 C March 8, 2010 26 / 39
Linear Algebra II: Vectors and Matrices Operations
Note:
How to understand this definition by using the definition of matrix-vectormultiplication?
We may think of each column of B as a column vector of length n.We know how to multiply each of these with the matrix A and thenwe put the resulting vectors into one matrix.
A =
26664a11 · · · a1n
a21 · · · a2n...
...am1 · · · amn
37775 , B =
26664b11 · · · b1k
b21 · · · b2k...
...bn1 · · · bnk
37775 =ˆ
B1 | B2 | · · · | Bk˜
where, B1 =
264 b11...
bn1
375, B2 =
264 b12...
bn2
375, . . . , B1 =
264 b1k...
bnk
375.
Then AB =ˆ
AB1 | AB2 | · · · | ABk˜
What is the size of matrix C, the product of A and B?
(m × n) ∗ (n× k) = (m× k).
Jing Li (UofO) MAT 1332 C March 8, 2010 27 / 39
Linear Algebra II: Vectors and Matrices Operations
Note:
How to understand this definition by using the definition of matrix-vectormultiplication?
We may think of each column of B as a column vector of length n.We know how to multiply each of these with the matrix A and thenwe put the resulting vectors into one matrix.
A =
26664a11 · · · a1n
a21 · · · a2n...
...am1 · · · amn
37775 , B =
26664b11 · · · b1k
b21 · · · b2k...
...bn1 · · · bnk
37775 =ˆ
B1 | B2 | · · · | Bk˜
where, B1 =
264 b11...
bn1
375, B2 =
264 b12...
bn2
375, . . . , B1 =
264 b1k...
bnk
375.
Then AB =ˆ
AB1 | AB2 | · · · | ABk˜
What is the size of matrix C, the product of A and B?
(m × n) ∗ (n× k) = (m× k).
Jing Li (UofO) MAT 1332 C March 8, 2010 27 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–
BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–
(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35
Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4: Consider the following matrices
A =
»1 23 4
–, B =
»1 2 3−1 0 4
–, C =
24 2 11 −1−1 0
35 , D =
»−1 −20 1
–Then, well-defined colored in blue, undefined colored in red.
AB =
»1 23 4
– »1 2 3−1 0 4
–=
»1 · 1 + 2 · (−1) 1 · 2 + 2 · 0 1 · 3 + 2 · 43 · 1 + 4 · (−1) 3 · 2 + 4 · 0 3 · 3 + 4 · 4
–=
»−1 2 9−1 6 25
–BA =
»1 2 3−1 0 4
– »1 23 4
–(2× 3) ∗ (2× 2) wrong.
BT A =
24 1 −12 03 4
35» 1 23 4
–
=
24 1 · 1 + (−1) · 3 1 · 2 + (−1) · 42 · 1 + 0 · 3 2 · 2 + 0 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
35 =
24 −2 −22 415 22
35Jing Li (UofO) MAT 1332 C March 8, 2010 28 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
CA =
24 2 11 −1−1 0
35» 1 23 4
–
=
24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4
35 =
24 5 8−2 −2−1 −2
35
ACT =
»1 23 4
– »2 1 −11 −1 0
–=
»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0
–=
»4 −1 −110 −1 −3
–
Jing Li (UofO) MAT 1332 C March 8, 2010 29 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
CA =
24 2 11 −1−1 0
35» 1 23 4
–
=
24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4
35 =
24 5 8−2 −2−1 −2
35
ACT =
»1 23 4
– »2 1 −11 −1 0
–=
»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0
–=
»4 −1 −110 −1 −3
–
Jing Li (UofO) MAT 1332 C March 8, 2010 29 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
CA =
24 2 11 −1−1 0
35» 1 23 4
–
=
24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4
35 =
24 5 8−2 −2−1 −2
35
ACT =
»1 23 4
– »2 1 −11 −1 0
–=
»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0
–=
»4 −1 −110 −1 −3
–
Jing Li (UofO) MAT 1332 C March 8, 2010 29 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
CA =
24 2 11 −1−1 0
35» 1 23 4
–
=
24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4
35 =
24 5 8−2 −2−1 −2
35
ACT =
»1 23 4
– »2 1 −11 −1 0
–=
»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0
–=
»4 −1 −110 −1 −3
–
Jing Li (UofO) MAT 1332 C March 8, 2010 29 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
CA =
24 2 11 −1−1 0
35» 1 23 4
–
=
24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4
35 =
24 5 8−2 −2−1 −2
35
ACT =
»1 23 4
– »2 1 −11 −1 0
–=
»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0
–=
»4 −1 −110 −1 −3
–
Jing Li (UofO) MAT 1332 C March 8, 2010 29 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
CA =
24 2 11 −1−1 0
35» 1 23 4
–
=
24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4
35 =
24 5 8−2 −2−1 −2
35
ACT =
»1 23 4
– »2 1 −11 −1 0
–=
»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0
–=
»4 −1 −110 −1 −3
–
Jing Li (UofO) MAT 1332 C March 8, 2010 29 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
CA =
24 2 11 −1−1 0
35» 1 23 4
–
=
24 2 · 1 + 1 · 3 2 · 2 + 1 · 41 · 1 + (−1) · 3 1 · 2 + (−1) · 4(−1) · 1 + 0 · 3 (−1) · 2 + 0 · 4
35 =
24 5 8−2 −2−1 −2
35
ACT =
»1 23 4
– »2 1 −11 −1 0
–=
»1 · 2 + 2 · 1 1 · 1 + 2 · (−1) 1 · (−1) + 2 · 03 · 2 + 4 · 1 3 · 1 + 4 · (−1) 3 · (−1) + 4 · 0
–=
»4 −1 −110 −1 −3
–
Jing Li (UofO) MAT 1332 C March 8, 2010 29 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–
=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–
=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–
=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 4 cont’d
A2 = AA =
»1 23 4
– »1 23 4
–=
»1 · 1 + 2 · 3 1 · 2 + 2 · 43 · 1 + 4 · 3 3 · 2 + 4 · 4
–=
»7 10
15 22
–
AD =
»1 23 4
– »−1 −20 1
–=
»1 · (−1) + 2 · 0 1 · (−2) + 2 · 13 · (−1) + 4 · 0 3 · (−2) + 4 · 1
–=
»−1 0−3 −2
–
DA =
»−1 −20 1
– »1 23 4
–=
»(−1) · 1 + (−2) · 3 (−1) · 2 + (−2) · 4
0 · 1 + 1 · 3 0 · 2 + 1 · 4
–=
»−7 −103 4
–
Jing Li (UofO) MAT 1332 C March 8, 2010 30 / 39
Linear Algebra II: Vectors and Matrices Operations
AD =
»−1 0−3 −2
–, DA =
»−7 −103 4
–,−→ AD 6= DA.
Note:
The order of the product matters!
Matrix multiplication is NOT commutative, even if both products are defined.
Jing Li (UofO) MAT 1332 C March 8, 2010 31 / 39
Linear Algebra II: Vectors and Matrices Operations
AD =
»−1 0−3 −2
–, DA =
»−7 −103 4
–,−→ AD 6= DA.
Note:
The order of the product matters!
Matrix multiplication is NOT commutative, even if both products are defined.
Jing Li (UofO) MAT 1332 C March 8, 2010 31 / 39
Linear Algebra II: Vectors and Matrices Operations
AD =
»−1 0−3 −2
–, DA =
»−7 −103 4
–,−→ AD 6= DA.
Note:
The order of the product matters!
Matrix multiplication is NOT commutative, even if both products are defined.
Jing Li (UofO) MAT 1332 C March 8, 2010 31 / 39
Linear Algebra II: Vectors and Matrices Operations
Multiplication of the identity matrix
For real number arithmetic, we know that 1 · a = a · 1 = a.
The identity matrix gives the same result in matrix multiplication
Multiplication of the identity matrix
If A is n ×m matrix, then we have
InA = AIm = A
Note: we really do need different identity matrices on each side of A that will dependupon the size of A.
Jing Li (UofO) MAT 1332 C March 8, 2010 32 / 39
Linear Algebra II: Vectors and Matrices Operations
Multiplication of the identity matrix
For real number arithmetic, we know that 1 · a = a · 1 = a.
The identity matrix gives the same result in matrix multiplication
Multiplication of the identity matrix
If A is n ×m matrix, then we have
InA = AIm = A
Note: we really do need different identity matrices on each side of A that will dependupon the size of A.
Jing Li (UofO) MAT 1332 C March 8, 2010 32 / 39
Linear Algebra II: Vectors and Matrices Operations
Multiplication of the identity matrix
For real number arithmetic, we know that 1 · a = a · 1 = a.
The identity matrix gives the same result in matrix multiplication
Multiplication of the identity matrix
If A is n ×m matrix, then we have
InA = AIm = A
Note: we really do need different identity matrices on each side of A that will dependupon the size of A.
Jing Li (UofO) MAT 1332 C March 8, 2010 32 / 39
Linear Algebra II: Vectors and Matrices Operations
Multiplication of the identity matrix
For real number arithmetic, we know that 1 · a = a · 1 = a.
The identity matrix gives the same result in matrix multiplication
Multiplication of the identity matrix
If A is n ×m matrix, then we have
InA = AIm = A
Note: we really do need different identity matrices on each side of A that will dependupon the size of A.
Jing Li (UofO) MAT 1332 C March 8, 2010 32 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 5:
For
A =
24 10 0−3 8−1 11
35 ,
then
I3A =
24 1 0 00 1 00 0 1
35 24 10 0−3 8−1 11
35 =
24 10 0−3 8−1 11
35AI2 =
24 10 0−3 8−1 11
35 » 1 00 1
–=
24 10 0−3 8−1 11
35Hence,
I3A = AI2 = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 33 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 5:
For
A =
24 10 0−3 8−1 11
35 ,
then
I3A =
24 1 0 00 1 00 0 1
35
24 10 0−3 8−1 11
35 =
24 10 0−3 8−1 11
35AI2 =
24 10 0−3 8−1 11
35 » 1 00 1
–=
24 10 0−3 8−1 11
35Hence,
I3A = AI2 = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 33 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 5:
For
A =
24 10 0−3 8−1 11
35 ,
then
I3A =
24 1 0 00 1 00 0 1
35 24 10 0−3 8−1 11
35
=
24 10 0−3 8−1 11
35AI2 =
24 10 0−3 8−1 11
35 » 1 00 1
–=
24 10 0−3 8−1 11
35Hence,
I3A = AI2 = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 33 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 5:
For
A =
24 10 0−3 8−1 11
35 ,
then
I3A =
24 1 0 00 1 00 0 1
35 24 10 0−3 8−1 11
35 =
24 10 0−3 8−1 11
35
AI2 =
24 10 0−3 8−1 11
35 » 1 00 1
–=
24 10 0−3 8−1 11
35Hence,
I3A = AI2 = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 33 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 5:
For
A =
24 10 0−3 8−1 11
35 ,
then
I3A =
24 1 0 00 1 00 0 1
35 24 10 0−3 8−1 11
35 =
24 10 0−3 8−1 11
35AI2 =
24 10 0−3 8−1 11
35
»1 00 1
–=
24 10 0−3 8−1 11
35Hence,
I3A = AI2 = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 33 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 5:
For
A =
24 10 0−3 8−1 11
35 ,
then
I3A =
24 1 0 00 1 00 0 1
35 24 10 0−3 8−1 11
35 =
24 10 0−3 8−1 11
35AI2 =
24 10 0−3 8−1 11
35 » 1 00 1
–
=
24 10 0−3 8−1 11
35Hence,
I3A = AI2 = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 33 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 5:
For
A =
24 10 0−3 8−1 11
35 ,
then
I3A =
24 1 0 00 1 00 0 1
35 24 10 0−3 8−1 11
35 =
24 10 0−3 8−1 11
35AI2 =
24 10 0−3 8−1 11
35 » 1 00 1
–=
24 10 0−3 8−1 11
35
Hence,I3A = AI2 = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 33 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 5:
For
A =
24 10 0−3 8−1 11
35 ,
then
I3A =
24 1 0 00 1 00 0 1
35 24 10 0−3 8−1 11
35 =
24 10 0−3 8−1 11
35AI2 =
24 10 0−3 8−1 11
35 » 1 00 1
–=
24 10 0−3 8−1 11
35Hence,
I3A = AI2 = A.
Jing Li (UofO) MAT 1332 C March 8, 2010 33 / 39
Linear Algebra II: Vectors and Matrices Operations
Operations of the zero matrix:
The zero matrix (denoted by 0 for a general matrix and 0 for a column/row vector) willtake the place of the number 0 in most of the matrix arithmetic.
Zero matrix properties: In the following properties, A is a matrix and 0 is the zeromatrix sized appropriately for the indicated operation to be valid.
1 A + 0 = 0 + A, (A is m × n, 0 is m × n)2 A− A = 0, (A is m × n, 0 is m × n)3 0− A = −A, (A is m × n, 0 is m × n)4 0A = 0, (A is m × n, the first 0 is k ×m, the second 0 is k × n)5 A0 = 0, (A is m × n, the first 0 is n × k , the second 0 is m × k )
Jing Li (UofO) MAT 1332 C March 8, 2010 34 / 39
Linear Algebra II: Vectors and Matrices Operations
Operations of the zero matrix:
The zero matrix (denoted by 0 for a general matrix and 0 for a column/row vector) willtake the place of the number 0 in most of the matrix arithmetic.Zero matrix properties: In the following properties, A is a matrix and 0 is the zeromatrix sized appropriately for the indicated operation to be valid.
1 A + 0 = 0 + A, (A is m × n, 0 is m × n)2 A− A = 0, (A is m × n, 0 is m × n)3 0− A = −A, (A is m × n, 0 is m × n)4 0A = 0, (A is m × n, the first 0 is k ×m, the second 0 is k × n)5 A0 = 0, (A is m × n, the first 0 is n × k , the second 0 is m × k )
Jing Li (UofO) MAT 1332 C March 8, 2010 34 / 39
Linear Algebra II: Vectors and Matrices Operations
Operations of the zero matrix:
Now, in real number arithmetic we know that
if ab = ac, and a 6= 0, then b = c. (cancellation law)
if ab = 0 then a = 0 and/or b = 0. (the zero factor property)
BUT, we see the following two examples:
Jing Li (UofO) MAT 1332 C March 8, 2010 35 / 39
Linear Algebra II: Vectors and Matrices Operations
Operations of the zero matrix:
Now, in real number arithmetic we know that
if ab = ac, and a 6= 0, then b = c. (cancellation law)
if ab = 0 then a = 0 and/or b = 0. (the zero factor property)
BUT, we see the following two examples:
Jing Li (UofO) MAT 1332 C March 8, 2010 35 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–
=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–
=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–
=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–
=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–
Note:AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C).
So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.
The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 6:
Consider
A =
»−3 2−6 4
–, B =
»−1 23 −2
–, C =
»1 46 1
–,
calculate AB and AC:
AB =
»−3 2−6 4
– »−1 23 −2
–=
»−3 · (−1) + 2 · 3 −3 · 2 + 2 · (−2)−6 · (−1) + 4 · 3 −6 · 2 + 4 · (−2)
–=
»9 −10
18 −20
–
AC =
»−3 2−6 4
– »1 46 1
–=
»−3 · 1 + 2 · 6 −3 · 4 + 2 · 1−6 · 1 + 4 · 6 −6 · 4 + 4 · 1
–=
»9 −1018 −20
–Note:
AB = AC , (A 6= 0, B 6= C). So from AB = AC we can’t have B = C.The cancellation law will not be valid in general for matrix multiplication.
Jing Li (UofO) MAT 1332 C March 8, 2010 36 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 7: (also see the last example in the additional lecture notes)
Consider
A =
»1 22 4
–, B =
»−16 2
8 −1
–,
Then
AB =
»1 22 4
– »−16 2
8 −1
–
=
»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)
–=
»0 00 0
–Note:
AB = 0 despite the face that A 6= 0, B 6= 0. In this case, the zero factor propertydose NOT hold.
There will be no zero factor property for the multiplication of any two randommatrices.
Jing Li (UofO) MAT 1332 C March 8, 2010 37 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 7: (also see the last example in the additional lecture notes)
Consider
A =
»1 22 4
–, B =
»−16 2
8 −1
–,
Then
AB =
»1 22 4
– »−16 2
8 −1
–=
»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)
–
=
»0 00 0
–Note:
AB = 0 despite the face that A 6= 0, B 6= 0. In this case, the zero factor propertydose NOT hold.
There will be no zero factor property for the multiplication of any two randommatrices.
Jing Li (UofO) MAT 1332 C March 8, 2010 37 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 7: (also see the last example in the additional lecture notes)
Consider
A =
»1 22 4
–, B =
»−16 2
8 −1
–,
Then
AB =
»1 22 4
– »−16 2
8 −1
–=
»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)
–=
»0 00 0
–
Note:
AB = 0 despite the face that A 6= 0, B 6= 0. In this case, the zero factor propertydose NOT hold.
There will be no zero factor property for the multiplication of any two randommatrices.
Jing Li (UofO) MAT 1332 C March 8, 2010 37 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 7: (also see the last example in the additional lecture notes)
Consider
A =
»1 22 4
–, B =
»−16 2
8 −1
–,
Then
AB =
»1 22 4
– »−16 2
8 −1
–=
»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)
–=
»0 00 0
–Note:
AB = 0 despite the face that A 6= 0, B 6= 0.
In this case, the zero factor propertydose NOT hold.
There will be no zero factor property for the multiplication of any two randommatrices.
Jing Li (UofO) MAT 1332 C March 8, 2010 37 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 7: (also see the last example in the additional lecture notes)
Consider
A =
»1 22 4
–, B =
»−16 2
8 −1
–,
Then
AB =
»1 22 4
– »−16 2
8 −1
–=
»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)
–=
»0 00 0
–Note:
AB = 0 despite the face that A 6= 0, B 6= 0. In this case, the zero factor propertydose NOT hold.
There will be no zero factor property for the multiplication of any two randommatrices.
Jing Li (UofO) MAT 1332 C March 8, 2010 37 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 7: (also see the last example in the additional lecture notes)
Consider
A =
»1 22 4
–, B =
»−16 2
8 −1
–,
Then
AB =
»1 22 4
– »−16 2
8 −1
–=
»1 · (−16) + 2 · 8 1 · 2 + 2 · (−1)2 · (−16) + 4 · 8 2 · 2 + 4 · (−1)
–=
»0 00 0
–Note:
AB = 0 despite the face that A 6= 0, B 6= 0. In this case, the zero factor propertydose NOT hold.
There will be no zero factor property for the multiplication of any two randommatrices.
Jing Li (UofO) MAT 1332 C March 8, 2010 37 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 8:
Consider A =
24 1−12
35 , B =ˆ
1 3 2˜, then AB =?.
Solutions:
A is 3× 1, and B is 1× 3. Then AB should be 3× 3.
AB =
24 1−12
35 ˆ 1 3 2˜
=
24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2
35=
24 1 3 2−1 −3 −22 6 4
35
Jing Li (UofO) MAT 1332 C March 8, 2010 38 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 8:
Consider A =
24 1−12
35 , B =ˆ
1 3 2˜, then AB =?.
Solutions: A is 3× 1, and B is 1× 3.
Then AB should be 3× 3.
AB =
24 1−12
35 ˆ 1 3 2˜
=
24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2
35=
24 1 3 2−1 −3 −22 6 4
35
Jing Li (UofO) MAT 1332 C March 8, 2010 38 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 8:
Consider A =
24 1−12
35 , B =ˆ
1 3 2˜, then AB =?.
Solutions: A is 3× 1, and B is 1× 3. Then AB should be 3× 3.
AB =
24 1−12
35 ˆ 1 3 2˜
=
24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2
35=
24 1 3 2−1 −3 −22 6 4
35
Jing Li (UofO) MAT 1332 C March 8, 2010 38 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 8:
Consider A =
24 1−12
35 , B =ˆ
1 3 2˜, then AB =?.
Solutions: A is 3× 1, and B is 1× 3. Then AB should be 3× 3.
AB =
24 1−12
35 ˆ 1 3 2˜
=
24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2
35=
24 1 3 2−1 −3 −22 6 4
35
Jing Li (UofO) MAT 1332 C March 8, 2010 38 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 8:
Consider A =
24 1−12
35 , B =ˆ
1 3 2˜, then AB =?.
Solutions: A is 3× 1, and B is 1× 3. Then AB should be 3× 3.
AB =
24 1−12
35 ˆ 1 3 2˜
=
24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2
35
=
24 1 3 2−1 −3 −22 6 4
35
Jing Li (UofO) MAT 1332 C March 8, 2010 38 / 39
Linear Algebra II: Vectors and Matrices Operations
Example 8:
Consider A =
24 1−12
35 , B =ˆ
1 3 2˜, then AB =?.
Solutions: A is 3× 1, and B is 1× 3. Then AB should be 3× 3.
AB =
24 1−12
35 ˆ 1 3 2˜
=
24 1 · 1 1 · 3 1 · 2−1 · 1 −1 · 3 −1 · 22 · 1 2 · 3 2 · 2
35=
24 1 3 2−1 −3 −22 6 4
35
Jing Li (UofO) MAT 1332 C March 8, 2010 38 / 39
Linear Algebra II: Vectors and Matrices Operations
Thank you!I will greatly appreciate it if you can tell me the typo that you find.
Jing Li (UofO) MAT 1332 C March 8, 2010 39 / 39