Transcript 4-5 * 2x2 Matrices, Determinants, & Inverses

4-5 – 2x2 Matrices,
Determinants, &
Inverses
Objectives
Evaluating Determinants of 2x2 Matrices
Using Inverse Matrices to Solve Equations
Vocabulary
A square matrix is a matrix with the same number of
columns as rows.
Multiplicative Identity Matrix is an n x n square matrix with
1’s along the main diagonal and 0’s elsewhere.
1 0
I2  

0
1


1 0 0 
I 3  0 1 0
0 0 1
Vocabulary
Multiplicative Inverse of a Matrix
If A and B are n x n matrices, and AB = BA = 1, then
B is the multiplicative inverse of A, written A-1
AA-1 = A-1 A = 1
Verifying Inverses
Show that matrices A and B are multiplicative inverses.
–1
1
3
7
A =
AB =
3
7
B =
–1
1
0.1
–0.7
0.1
–0.7
0.1
0.3
0.1
0.3
=
(3)(0.1) + (–1)(–0.7)
(7)(0.1) + (1)(–0.7)
=
1
0
(3)(0.1) + (–1)(0.3)
(7)(0.1) + (1)(0.3)
0
1
AB = I, so B is the multiplicative inverse of A.
Vocabulary
The determinant of a 2 x 2 matrix
det A
a b 
c d 


a b
c d
is ad – bc.
Evaluating the Determinant of a
2x2 Matrix
Evaluate each determinant.
a. det
7
–5
8
–9
7
–5
b. det
4
5
–3
6
=
4
5
–3
6
= (4)(6) – (–3)(5) = 39
c. det
a
b
–b
a
=
a
b
–b
a
= (a)(a) – (–b)(b) = a2 + b2
=
8
–9
= (7)(–9) – (8)(–5) = –23
Vocabulary
a b 
Let A = 
 . If det A = 0, then A has no inverse.
c
d


1  d  b
1  d  b

If det A ≠ 0, then A =


det A  c a  ad  bc  c a 
Finding an Inverse Matrix
Determine whether each matrix has an inverse. If it does,
find it.
a.
X =
12
9
4
3
ad – bc = (12)(3) – (4)(9)
Find det X.
Simplify.
= 0
Since det X = 0, the inverse of X does not exist.
b.
Y =
6
25
5
20
ad – bc = (6)(20) – (5)(25)
Find det Y.
Simplify.
= –5
Since the determinant =/ 0, the inverse of Y exists.
Continued
(continued)
Y–1
1
=
det Y
20
–25
–5
6
1
det Y
20
–25
–5
6
=
= – 1
5
=
–4
5
20
–25
1
–1.2
–5
6
Change signs.
Switch positions.
Use the determinant to
write the inverse.
Substitute –5 for the
determinant.
Multiply.
Solving a Matrix Equation
Solve
9
4
25
11
X =
3
–7
for the matrix X.
The matrix equation has the form AX = B. First find A–1.
A–1
1
=
ad – bc
d
–c
–b
a
1
=
(9)(11) – (25)(4)
=
–11
4
Use the definition
of inverse.
11
–4
25
–9
–25
9
Substitute.
Simplify.
Use the equation X = A–1B.
X =
–11
4
25
–9
3
–7
Substitute.
Continued
(continued)
=
(–11)(3) + (25)(–7)
(4)(3) + (–9)(–7)
Check:
–208
75
=
Multiply and
simplify.
3
–7
Use the original
equation.
–208
75
3
–7
Substitute.
9(–208) + 25(75)
4(–208) + 11(75)
3
–7
Multiply and simplify.
3
–7
3
–7
9
4
9
4
25
11
25
11
X =
=
Real World Example
In a city with a stable group of 45,000 households, 25,000
households use long distance carrier A, and 20,000 use long distance
carrier B. Records show that over a 1-year period, 84% of the
households remain with carrier A, while 16% switch to B. 93% of the
households using B stay with B, while 7% switch to A.
a. Write a matrix to represent the changes in long distance carriers.
From
To A
To B
A
0.84
0.16
B
0.07
0.93
Write the percents as decimals.
Continued
(continued)
b. Predict the number of households that will be using distance carrier B
next year.
Use A
Use B
0.84
0.16
25000
20000
0.07
0.93
Write the information in a matrix.
25000
20000
=
22,400
22,600
22,600 households will use carrier B.
Continued
(continued)
c. Use the inverse of the matrix from part (a) to find, to the nearest
hundred households, the number of households that used carrier A
last year.
First find the determinant of
0.84
0.16
0.07
0.93
0.84
0.16
0.07
.
0.93
= 0.77
Multiply the inverse matrix by the information matrix in part (b).
Use a calculator and the exact inverse.
1
0.77
0.93
–.016
–0.07
0.84
25,000
20,000
28,377
16,623
About 28,400 households used carrier A.
Homework
Pg 203 # 1,4,5,14,18, 22