Download 4.4 Lesson

HW: Pg. 219 #16-26e, 31, 33
HW: Pg. 219-220 #37, 41, 45, 49, 59
HW: Pg. 219-220 #37, 41, 45, 49, 59
HW: Pg. 219-220 #37, 41, 45, 49, 59
HW: Quiz 1 Pg. 221 #7-17o
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
3 8 
A

2
5


5 -8
1
1 5 -8 1 5 -8
A 

 




3(15)-2(8) -2 3 15  16 -2 3 1 -2 3
5 -8  5 +8 
 1 



-2
3
+
2
3

 

1
 5
so A  
 +2
1
+8 
- 3
for Example 1
GUIDED PRACTICE
Find the inverse of the matrix.
1.
6 1
2 4
ANSWER
2.
–1 5
3.
–3 –4
–4 8
–1 –2
ANSWER
ANSWER
2
11
– 1
22
2
3
– 5
12
–1
2
– 1
11
3
11
1
3
– 1
12
1
2
–3
2
EXAMPLE 2
Solve a matrix equation
Solve the matrix equation AX = B for the 2 × 2 matrix X.
A
2
–7
B
X =
–1 4
–21
3
12
–2
SOLUTION
Begin by finding the inverse of A.
A–1
=
1
8–7
4
7
1
2
=
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.
4 7
1 2
2
–1
–7
4
1 0
X
=
X =
0 1
X=
4 7
–21
3
1 2
12
–2
0 –2
A–1 AX = A–1 B
IX = A–1 B
3 –1
0 –2
3 –1
X = A–1 B
GUIDED PRACTICE
for Example 2
4. Solve the matrix equation
–4 1
0 6
ANSWER
–1
–2
4
1
X=
8
9
24 6
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.
A=
2
1 –2
5
3
0
4
3
8
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.
for Example 3
GUIDED PRACTICE
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.
5.
A=
2
–2
0
2
0
–2
12 –4
–6
for Example 3
GUIDED PRACTICE
–3 4 5
6.
A=
1 5 0
5 2 2
7.
2 1 –2
A=
5 3
0
4 3
8
EXAMPLE 4
Solve a linear system
Use an inverse matrix to solve the linear system.
2x – 3y = 19
Equation 1
x + 4y = –7
Equation 2
SOLUTION
STEP 1
Write the linear system as a matrix equation AX = B.
coefficient
matrix (A)
(X)
2 –3
.
1 4
matrix of
matrix of
variables
constants(B)
x
y
=
19
–7
EXAMPLE 4
STEP 2
Solve a linear system
Find the inverse of matrix A.
1
8 – (–3)
A–1 =
–1 2
=
3
11
– 1
11
2
11
Multiply the matrix of constants by A–1 on the left.
STEP 3
X=
4 3
4
11
A–1B
=
4
11
3
11
– 1
11
2
11
19
–7
ANSWER
The solution of the system is (5, – 3).
CHECK
2(5) – 3(–3) = 10 + 9 = 19
5 + 4(–3) = 5 – 12 = –7
=
5
–3
x
=
y
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.
EXAMPLE 5
Solve a multi-step problem
SOLUTION
STEP 1
Write verbal models for the situation.
EXAMPLE 5
STEP 2
Solve a multi-step problem
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
STEP 3
Equation 1
2m + 2p + d = 37.00
Equation 2
4m + 3p + 2d = 72.50
Equation 3
Rewrite the system as a matrix equation.
2 1 0
m
2 2 1
p
4 3 2
d
15.50
=
37.00
72.50
EXAMPLE 5
STEP 4
Solve a multi-step problem
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.
2x – y = – 6
9.
8.
4x + y = 10
6x – 3y = – 18
3x + 5y = –1
ANSWER
ANSWER
(3, –2)
11.
infinitely many solutions
10.
3x – y = –5
–4x + 2y = 8
ANSWER
(– 1, 2)
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?
ANSWER
movie pass: $8
package of popcorn: $1
DVD: $17
Homework:
Pg. 227 #17-23o, 28, 30, 31, 33-36