Chapter 2 Part 2 MATRICES - avon-schools.org · Final solution = numbers in the “answer” column...

Finite Math B Chapter 2 MATRICES 1

Chapter 2 – Part 2 MATRICES

A: Augmented Matrices and Row Operations (Lessons 2.2 pg 68 - 70)

Augmented Matrices Suppose you are given a system of The system can be written as a matrix: equations such as:

2 2

3 2 1

2

x y z

x y z

x y z

2 1 1 2

1 3 2 1

1 1 1 2

To separate the coefficients of the variables from the constants after the equals signs, we draw in a vertical line in the matrix. This is called an augmented matrix. Example 1: Write an augmented matrix for each system of equations. Do not solve.

a) 3 4 2

8 4

x y

x y

b)

5 25

4 10

x

x y

c)

2 3 5 12

3 4 15

5

x y z

x z

x y z

Example 2: Write the system of equations associated with each augmented matrix.

a) 1 3 6

2 3 10

b)

3 1 0 5

2 1 9 10

0 4 1 8


(example 2 continued…. Write the system of equations associated with each augmented matrix)

c)

1 0 0 3

0 1 0 5

0 0 1 4

d) 1 0 4

0 1 5

In algebra, our goal when faced with a system of equations is to find a solution for x, y, and z that makes the system of equations true.

Note that when you see this pattern: 1

2

3

1 0 0 #

0 1 0 #

0 0 1 #

You end up with

1

2

3

#

#

#

x

y

z

(1’s on main diagonal, 0’s elsewhere, constants last column)

Row Operations We will be using “Row Operations” to manipulate matrices to help us solve systems of equations. Example 3: Use the indicated row operations to change each matrix. a) Interchange R1 with R2.

What you are allowed to do: 1. INTERCHANGE TWO ROWS 2. MULTIPLY THE ELEMENTS OF A ROW BY A NONZERO REAL NUMBER 3. ADD A NONZERO MULTIPLE OF THE ELEMENTS OF ONE ROW TO THE CORRESPONDING ELEMENTS OF A NONZERO MULTIPLE OF SOME OTHER ROW.

0 4 2 1

1 1 1 2

6 6 0 9


(Example 3 continued: Use the indicated row operation to change each matrix) b) Replace R3 by 1

3R3 .

c) Replace R2 with (-2)R1 +R2

3 1 4 1

6 5 0 2

1 3 4 5

d) Replace R3 with (-3)R2 + 5R3

3 1 4 1

0 5 0 2

0 3 4 5

e) Replace R1 with (-2)R3 + 3R1

1 0 4 7

0 5 7 2

0 0 6 24

f) Replace R2 with (-7)R3 + 6R2

1 0 4 7

0 5 7 2

0 0 6 24

g) Replace R2 with (-1)R1 +4R2

4 3 3 7

1 1 7 2

6 4 6 2

1 3 2 1

0 1 1 2

6 6 3 9


B: Gauss-Jordan Method for Solving Systems of Equations (Lesson 2.2 , textbook pg 70 – 80)

Problem: Find the solution to a system of equations like

5 6

3 3 10

3 2 5

x y z

x y z

x y z

Strategy: 1. Write the system of equations as an augmented matrix 2. Use Row Operations to transform the matrix into a matrix with whole numbers on the main diagonal, but 0’s elsewhere. 3. Use Row Operations to transform the matrix into an “identity” matrix. (1’s on diagonal, 0’s elsewhere) 4. Final solution = numbers in the “answer” column of the matrix.

Example: 15

31 0

0 1

Example:

1 0 0 4

0 1 0 2

0 0 1 3

Meaning: Meaning:

Make it happen: GAUSS-JORDAN Your Goal:

1 0

0 1

#

#

or

1 0 0

0 1 0

0 0 1

#

#

#

Using legal row operations: 2x2 System Clear Col 1 Clear Col 2 Create 1’s Main Diag.

0

# # #

# #

0

0

# #

# #

1 0

0 1

#

#

3x3 System Clear Col 1 Clear Col 2 Clear Col 3 Create 1’s Main Diag.

0

0

# # # #

# # #

# # #

0

0

0 0

# # #

# # #

# #

0 0

0 0

0 0

# #

# #

# #

1 0 0

0 1 0

0 0 1

#

#

#


Example 1 : Use the Gauss-Jordan Method to solve each system of equations

a) 2 4 2

3 5 0

x y

x y

b) 3 4 1

5 2 19

x y

x y

c) 2 2

3 6 5

x y

x y


(Example 1 continued): Use the Gauss-Jordan Method to solve each system of equations

d)

3

2 3 7 0

3 2 17

x y z

x y z

x y z


(Example 1 Continued): Use the Gauss-Jordan Method to solve each system of equations

e)

2 5 4 8

2 2 4

2 2

x y z

x z

x y z


(Example 1 Continued): Use the Gauss-Jordan Method to solve each system of equations

f)

5 6

3 3 10

3 2 5

x y z

x y z

x y z


C: Matrix Inverses (Lessons 2.5 pg 107-111)

Additive Inverse Vs Multiplicative Inverse For numbers, algebraic expressions, and matrices “zero” or the “zero matrix” plays the role of the Additive Identity. The sum of a value and the additive identity is the original value, unchanged.

144 0 3 0x 3 5 0 0

6 2 0 0

If A and B are additive inverses then their sum gives you the additive inverse value, then A + B = 0 Examples of Additive Inverses:

144 + = 0 -3x + = 0 3 5

6 2

+ = 0 0

0 0

For numbers and algebraic expressions, 1 plays the role of the Multiplicative Identity. For matrices, the “Identity Matrix” is the Multiplicative Identity. The product of the a value and the multiplicative identity is the original value, unchanged.

144 x 1 = 3 1 =x 3 5 1 0

6 2 0 1

If A and B are Multiplicative Inverses then their product gives you the multiplicative inverse value, then A x B = 1 Examples of Multiplicative Inverses:

144 x = 1 3 = 1x 3 5

6 2

???? =

1 0

0 1

Finding a multiplicative inverse for a matrix is more challenging than just reciprocating each value in the matrix due to the unique ROWS x COLUMNS requirement of multiplying matrices.


Terminology/Facts

Only a square matrix can have a multiplicative inverse.

If A is your given matrix, then 1A is the multiplicative matrix.

I is the identity matrix.

The identity matrix I is a square matrix with 1’s on the main diagonal and 0’s elsewhere

1A A I A I A

Example 1: Decide whether the given matrices are inverses of each other. (Check to see if their product is the identity matrix I )

a) 2 1 3 1

5 3 5 2and

b)

2 6 1 2

2 4 2 4and

c)

1 3 3 7 3 3

1 4 3 1 1 0

1 3 4 1 0 1

and

Finding the Multiplicative Inverse of a Matrix

Step 1: Form an augmented matrix that looks like A I

Step 2: Perform row operations to get the matrix in the form I B if possible.

Step 3: Matrix B is 1A


Example 2: Find the multiplicative inverse of each matrix

a) 1 1

2 0

b) 3 1

5 2

c) 1 3

2 6


(Example 2 continued) : Find the multiplicative inverse of each matrix

d)

1 1 1

4 5 0

0 1 3

e)

1 2 3

3 2 1

1 0 1


D: Using Matrix Inverses to Solve Systems of Equations (Lessons 2.5 pg 111-115) Any system of equations can be written as AX B where A the square coefficient matrix X the column matrix of the variables B the matrix of constants

Example: 2 5 15

4 9

x y

x y

Solving A System of Equations Using the Inverse Method

1. Find the 1A , the inverse of the coefficient matrix

2. Multiply 1A B 3. Check your answers Example 1: Solve each system of equations by using the inverse of the coefficient matrix.

a) 2 5 15

4 9

x y

x y


(Example 1 Continued): Solve each system of equations by using the inverse of the coefficient matrix.

b) 2 7 14

3 4 8

x y

x y

c)

1

4 5 2

3 3

x y z

x y

y z

We already found the inverse of this matrix in exercise 2d on page 12. Go back and copy that answer here:


(Example 1 continued): Solve each system of equations by using the inverse of the coefficient matrix.

d)

2 1

5

8

x z

y z

x y

Chapter 2 Part 2 MATRICES - avon-schools.org · Final solution = numbers in the “answer” column...

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