HW: Pg. 219 #16-26e, 31, 33. HW: Pg. 219-220 #37, 41, 45, 49, 59.

HW: Pg. 219 #16-26e, 31, 33

HW: Pg. 219-220 #37, 41, 45, 49, 59

HW: Quiz 1 Pg. 221 #7-17o

4.4 Identity and Inverse Matrices

EXAMPLE 1 Find the inverse of a 2 X 2 matrix

Find the inverse of A = .3 8

2 5

1

1

8

2

-8 -8 -8

-2 -2 -2

5

5

3

3 3 3

5 5

5 5

5

3 3

3

1 1 1

3(15)-2(8) 15 16 1

+1

-8 8

-2 2

+

8

2

-

+

+ -

A

A

A

so

GUIDED PRACTICE for Example 1

Find the inverse of the matrix.

1. 6 1

2 4

2

11

1

11

3

11–

1

22–

ANSWER

2. –1 5

–4 8

2

3

1

3

5

12–

1

12–

ANSWER

3. –3 –4

–1 –2

–1

1

2

2

3

2

–

ANSWER

EXAMPLE 2 Solve a matrix equation

SOLUTION

Begin by finding the inverse of A.

4 7

1 2=

Solve the matrix equation AX = B for the 2 × 2 matrix X.

2 –7

–1 4

–21 3

12 –2

A B

X =

A–1 =1

8 – 7

4 7

1 2

EXAMPLE 2 Solve a matrix equation

To solve the equation for X, multiply both sides of the equation by A– 1 on the left.

A–1 AX = A–1 B

IX = A–1 B

X = A–1 BX =0 –2

3 –1

4 7

1 2

–21 3

12 –2=

2 –7

–1 4

4 7

1 2X

X 1 0

0 1

0 –2

3 –1=


4. Solve the matrix equation –4 1

0 6X =

8 9

24 6

–1 –2

4 1

ANSWER

EXAMPLE 3 Find the inverse of a 3 × 3 matrix

Use a graphing calculator to find the inverse of A. Then use the calculator to verify your result.

2 1 – 2

5 3 0

4 3 8

A =

SOLUTION

Enter matrix A into a graphing calculator and calculate A–1. Then compute AA–1and A–1A to verify that you obtain the 3 × 3 identity matrix.


5. 2 –2 0

2 0 –2

12 –4 –6A =

Use a graphing calculator to find the inverse of the matrix A. Check the result by showing that AA-1= I and A-1A = I.


6. – 3 4 5

1 5 0

5 2 2

A =

7. 2 1 – 2

5 3 0

4 3 8

A =

EXAMPLE 4 Solve a linear system

Use an inverse matrix to solve the linear system.

2x – 3y = 19

x + 4y = –7

Equation 1

Equation 2

SOLUTION

STEP 1 Write the linear system as a matrix equation AX = B.

coefficient matrix of matrix of

matrix (A) (X) variables constants(B)

2 –3

1 4

. x

y

19

–7=

EXAMPLE 4 Solve a linear system

STEP 2 Find the inverse of matrix A.

4 3

–1 2

=A–1 =1

8 – (–3)

4

11

1

11

3

11

2

11–

STEP 3 Multiply the matrix of constants by A–1 on the left.

X = A–1B =

4

11

1

11

3

11

11– 2

19

–7=

5

–3=

x

y

The solution of the system is (5, – 3).

ANSWER

CHECK 2(5) – 3(–3) = 10 + 9 = 19

5 + 4(–3) = 5 – 12 = –7

EXAMPLE 5 Solve a multi-step problem

Gifts

A company sells three types of movie gift baskets. A basic basket with 2 movie passes and 1 package of microwave popcorn costs $15.50. A medium basket with 2 movie passes, 2 packages of popcorn, and 1 DVD costs $37. A super basket with 4 movie passes, 3 packages of popcorn, and 2 DVDs costs $72.50. Find the cost of each item in the gift baskets.


SOLUTION

STEP 1 Write verbal models for the situation.


STEP 2 Write a system of equations. Let m be the cost of a movie pass, p be the cost of a package of popcorn, and d be the cost of a DVD.

2m + p = 15.50 Equation 1

2m + 2p + d = 37.00 Equation 2

4m + 3p + 2d = 72.50 Equation 3

STEP 3 Rewrite the system as a matrix equation.

2 1 0

2 2 1

4 3 2

m

p

d

15.50

37.00

72.50

=


STEP 4 Enter the coefficient matrix A and the matrix of constants B into a graphing calculator. Then find the solution X = A–1B.

A movie pass costs $7, a package of popcorn costs $1.50, and a DVD costs $20.

GUIDED PRACTICE for Examples 4 and 5

Use an inverse matrix to solve the linear system.

4x + y = 10

3x + 5y = –1

8.

(3, –2)

ANSWER

9. 2x – y = – 6

6x – 3y = – 18

infinitely many solutions

ANSWER

10. 3x – y = –5

–4x + 2y = 8

(– 1, 2)

ANSWER

11. What if? In Example 5, how does the answer change if a basic basket costs $17, a medium basket costs $35, and a super basket costs $69?

movie pass: $8package of popcorn: $1

DVD: $17

ANSWER

Homework:

Pg. 227 #17-23o, 28, 30, 31, 33-36

HW: Pg. 219 #16-26e, 31, 33. HW: Pg. 219-220 #37, 41, 45, 49, 59.

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Transcript of HW: Pg. 219 #16-26e, 31, 33. HW: Pg. 219-220 #37, 41, 45, 49, 59.