Algebra 2 Chapter 4 Notes Matrices & Determinants Algebra 2 Chapter 4 Notes Matrices & Determinants.

Algebra 2

Chapter 4 Notes

Matrices & Determinants

a MATRIX is a rectangular arrangement of rows & columns

DIMENSIONS are the number of rows (horizontal) by the number of columns (vertical)

Matrix A = This matrix has 2 rows x 3 columns, so it is a 2 x 3 matrix.

6 2 ─1

─2 0 5

To add or subtract matrices, just add or subtract the corresponding entries.NOTE: you can add or subtract only if each matrix has the same dimensions.

3

─4

7

1

0

3+ =

4

─4

10

8 3

4 0

2 ─7

6 ─1- =6 10

─2 1

REMEMBER: “Rows x Columns” when describing a Matrix.

Matrix Operations 4.1

Multiplying Scalars

In matrix algebra, a real number is often called a scalar

To multiply a matrix by a scalar, you multiply each entry in the matrix by the scalar. This process is called “Scalar Multiplication.””

+

=─2 0

4 7

─6 0

12 21

=

3

1 ─2

0 3

─4 5

─2─4 5

0 ─8

─2 6

─2 4

0 ─6

8─10

+─4 5

0 ─8

─2 6

=─6 9

6 ─14

6 ─4

4.1Multiplying Scalars

REMEMBER: “Rows x Columns” when describing a Matrix.It is NOT “ Columns x Rows”

Properties of Matrix Operations

Associative Property of + ( A + B ) + C = A + ( B + C )

Commutative Property of + A + B = B + A

Multiplication of a sum or difference by a scalar obeys the distributive property

Distributive Property of + C ( A + B ) = CA + CB

Distributive Property of ─ C ( A ─ B ) = CA ─ CB

4.2Properties of Matrix Operations

The product of 2 matrices A and B is defined provided the number of columns in matrix A = the number of rows in matrix B

A x B = AB

m x n n x p m x p

dimensions of AB

Example 1: 3 x 2 x 2 x 4 = 3 x 4

6 entries 8 entries 12 entries

1 2 3 4

5 6 7 8

9 10 11 12

1 2

3 4

5 6

1 2 3 4

5 6 7 8

Multiplying Matrices 4.2

To multiply each matrix, multiply EACH Row by EACH column

Properties of Matrix Mulltiplication

Associative Property of x ( A B ) C = A ( B C )

Left distributive A (B + C) = AB + AC

Right distributive ( A + B ) C = AC + BC

Associative Property of a scalar C ( A B ) = (CA) B = A (CB)

Properties of Matrix Multiplications 4.2

1 2

3 4

5 6

•7 8

9 10 =

1 x 7 + 2 x 9 1 x 8 + 2 x 10

3 x 7 + 4 x 9 3 x 8 + 4 x 10

5 x 7 + 6 x 9 5 x 8 + 6 x 10

=7 + 18 = 25 8 + 20 = 28

21 + 36 = 57 24 + 40 = 64

35 + 54 = 89 40 + 60 = 100

=25 28

57 64

89 100

A B AB 3 x 2 x 2 x 2 = 3 x 2

REMEMBER: you multiply each ENTIRE row by each ENTIRE column. You multiple and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix.


1 2

3 4

5 6

•7 8

9 10 =

1 x 7 + 2 x 9 1 x 8 + 2 x 10

3 x 7 + 4 x 9 3 x 8 + 4 x 10

5 x 7 + 6 x 9 5 x 8 + 6 x 10

=7 + 18 = 25 8 + 20 = 28

21 + 36 = 57 24 + 40 = 64

35 + 54 = 89 40 + 60 = 100

=25 28

57 64

89 100

A B AB 3 x 2 x 2 x 2 = 3 x 2

REMEMBER: you multiply each ENTIRE row by each ENTIRE column and then add the result. That is why the number of columns of the first matrix must equal the number of rows of the second matrix.


─2 3

1─4

6 0

•─1 3

─2 4 =

─2 x ─1 + 3 x ─2

─2 x 3 + 3 x 4

1 x ─1 + ─ 4 x ─2

1 x 3 + ─ 4 x 4

6 x ─1 + 0 x ─2

6 x 3 + 0 x 4

=2 + ─ 6 = ─ 4 ─ 6 + 12 = 6

─ 1 + 8 = 7 3 ─ 16 = ─13

─ 6 + 0 = ─ 6 18 + 0 = 18

=

─ 4 6

7 ─13

─ 6 18

A B AB 3 x 2 x 2 x 2 = 3 x 2

REMEMBER: you multiply each row by each column. That is why the number of columns of the first matrix must equal the number of rows of the second matrix.


deta b

c d =

Determinants and Cramer’s RuleAssociated with each square matrix is a real # called it’s determinant, denoted by det A or │ A │.

a b

c d = ad ─ bc

The determinant of a 2 x 2 matrix is the difference of the products of the entries on the diagonals

det

Example 1

1 3

2 5 =1 3

2 5 = 1 • 5 ─ 2 • 3 = 5 ─ 6 = ─ 1

Determinants and Cramer’s Rule 4.3

det =

2 steps to get the determinant of a 3 x 3 matrix:1. Repeat first 2 columns to the right of the matrix2. Select the sum of the products the products in red from the sum of the

products in blue.

=

det

Example 1

= =

a b c

d e f

g h i

a b c

d e f

g h i

a b

d e

g h

(aei + bfg + cdh) ─ (gec + hfa + idb)

2 ─ 1 3

─ 2 0 f

1 2 4

2 ─ 1 3

─ 2 0 f

1 2 4

2 ─ 1

─ 2 0

1 2

(0 + ─ 1 + ─ 12) ─ (0 + 4 + 8) = ─ 13 ─ 12 = ─ 25


A = ± ½

A Determinant can find the area of a triangle whose vertices are points in a coordinate plane. Area of a Triangle with vertices , ( x1 , y1 ) , ( x2 , y2 ) , ( x3 , y3 )

A = ± ½

Example 1

= ± ½

x1 y1 1

x2 y2 1

x3 y3 1

1 2 1

4 0 1

6 2 1

(0 + 12 + 8) ─ (0 + 2 + 8) = ½ (20 ─ 10) = 5

Where ± indicates the appropriate sign should be chosen to yield a positive sign.

•

( x1 , y1 )

( x2 , y2 ) ( x3 , y3 )

1 2

4 0

6 2

•

•


Cramer’s Rule for a 2 x 2 matrix: Let A be the coefficient matrix of this linear system.

Linear System:

ax + by = ecx + dy = f

Coefficient Matrix:

a b

c d = det A

If det A ≠ 0, then the system has exactly one solution, the solution is:

e b

f dx =

det A

a e

c fy =and

det A



8 x + 5 y = 22 x ─ 4 y = ─ 10

Coefficient Matrix:

8 5

2─ 4

=

det A

If det A ≠ 0, then the system has exactly one solution, the solution is:

e b

f dx =

det A

a e

c fy =

det A

Linear System:

a x + b y = ec x + d y = f

=

(8) (─ 4 ) ─ (2) (5 ) (─ 32 ) ─ (10)

(─

42 )

=2 5

─ 10 ─ 4

─ 42

=8 2

2 ─ 10

─ 42

(2) (─ 4 ) ─ (─ 10) (5 ) = (─ 8) + (50) = = 42 ─ 42 ─ 42 ─ 42

= (8) (─ 10 ) ─ (2) (2 ) ─ 42

=

= ─ 1

(─ 80) ─ (4) ─ 42

= ─ 84 ─ 42

= 2

( x , y ) (─ 1 , 2 )



Coefficient Matrix:

=det A

x =

det A

y =

det A

Linear System:

a x + b y + c z = jd x + e y + f z = kg x + h y + i z = l

( x , y , z )

a b c

d e f

g h i

a b

d e

g h

= ( aei +bfg + cdh ) – ( ceg + afh + bdi )

If det A ≠ 0, then the system has exactly one solution, the solution is:j b c

k e f

l H i

a j c

d k f

g l i

z =

a b j

d e k

g H l

det A

4.3


Coefficient Matrix:

=det A

x =

– 10

y =

– 10

Linear System:

x + 4 y = 163 x + 8 y + 3 z = 92 2 y + z = 18

( x , y , z )( 12 , 1 , 16 )

1 4 0

3 8 3

0 2 I

1 4

3 8

0 2

= ( 8 + 0 + 0 ) – ( 0 + 6 + 12 ) = – 10

If det A ≠ 0, then the system has exactly one solution, the solution is:16 4 0

92 8 3

18 2 1

1 16 0

3 92 3

0 18 1

z =

1 4 16

3 8 92

0 2 18

16 4

92 8

18 2

1 16

3 92

0 18

1 4

3 8

0 2

= ( 128 + 216 + 0 ) – ( 0 + 96 + 368) – 10

= ( 92 + 0 + 0 ) – ( 0 + 54 + 48) – 10

= ( 144 + 0 + 96 ) – ( 0 + 184 + 216) – 10

= – 120 – 10

= – 10 – 10

= – 160 – 10

= 12

= 1

= 16

– 10

4.3

Identity and Inverse MatricesThe number 1 is the multiplicative identity for real #’s because 1 • a = a • 1 = aFor matrices, the n x n Identity matrix is the matrix that has 1’s on the main diagonal and 0’s elsewhere.

I =

1 0 0

0 1 0

0 0 I

1 0

0 1I =

2 x 2 Identity matrix 3 x 3 Identity matrix

If A is any n x n matrix and I is the Identity matrix, then I A = A and A I = A

Two n x n matrices are inverses of each other if their products (in both orders) is the n x n identity matrix. Remember: 3/4 • 4/3 = 1

AB =

BA =

3 –1

–5 2

2 1

5 3

2 1

5 3

3 –1

–5 2

1 0

0 1

1 0

0 1

=

=

= I

= I

Identity and Inverse Matricies 4.4

Inverse Matrices

The inverse of a 2 x 2 matrix A = is a b

c d

A –1 = 1 │ A │

d–b

–c a

= 1ad–cb provided ad–cb ≠ 0d

–b

–c a

13•2–4•1

A –1 =

2–1

–4

3

Remember : │ A │ and ad–cb are the determinant of matrix A.

3 1

4 2A =

Example:

= 1 6–4

2–1

–4

3

= 1 2

2–1

–4

3

=1

–12

–232

Identity and Inverse Matricies 4.4

Inverse Matrices a b

c d

A –1 = 1ad–cb

d –b–c a

A =

Switch a and d Change b and c to opposite signs Multiply by 1 divided by determinant of matrix A

4.4

Example 2: Solve matrix equation,

A X = B for 2 x 2 matrix X4 –1

–3 1

8 –5

–6 3

X =

Solution: Find inverse of A

A –1 = 1 4 – 3

1 1

3 4=

1 1

3 4

4 –1

–3 1

1 1

3 4 X = 8 –5

–6 3

1 1

3 4

1 0

0 1 X = 2 –2

0 –3

X = 2 –2

0 –3

A B

A X = B1 A X = 1 B A A 1 X = B

A X = B

A

Solving a Matrix Equation 4.4

Solving systems using inverse matricies

5 –4

1 2

Remember: Row x Column=

Coefficient matrix

Matrix of variables

8

6

x

y

Matrix of constants

5 x – 4 y = 8

1 x + 2 y = 6•

Example 1: Write a system of linear equations as a matrix equation

– 3 x + 4 y = 5

2 x – 1 y = – 10

–3 4

2 – 1

x

y =5

– 10

Example 2: Use matricies to solve the linear system

A –1 = 1 3 – 8

– 1 –4

–2 –3

=

15

45

25

35

A –1 = 1 – 5

– 1 –4

–2 –3

x

y =5

– 10

15

45

25

35

=1 ─ 8

2 – 10

= ─ 7

– 4

( X , Y ) (─ 7 , – 4 )

Solving Systems using Inverse Matricies 4.4

6 2 ─1

─2 0 5

8 3

4 ─1

3

─2

7

1 4 ─2

0 0 0

0 0 0

1 4

0 ─2

0 0

6 2 ─1

1 2 3 4

5 6 7 8

9 10 11 12

1 2 3 4

5 6 7 8

1

0

Algebra 2 Chapter 4 Notes Matrices & Determinants Algebra 2 Chapter 4 Notes Matrices & Determinants.

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