A First Course in Linear Algebra, 2017a

78 Matrices
What if the right side, B, of2.10 had been ⎣
⎡
⎣
⎡
0
1
3
1 0 1
1 −1 1
1 1 −1
⎤
⎦? In other words, what would be the solution to
⎤⎡
⎦⎣
x
y
z
⎤
⎡
⎦ = ⎣
0
1
3
⎤
⎦?
By the above discussion, the solution is given by
⎡
⎡ ⎤
1 1
0
x
2 2
⎣ y ⎦ = A −1 B = ⎢
⎣ 1 −1 0
z
1 − 1 2
− 1 2
⎤
⎡
⎥⎣
⎦
0
1
3
⎤
⎡
⎦ = ⎣
2
−1
−2
⎤
⎦
This illustrates that for a system AX = B where A −1 exists, it is easy to find the solution when the vector
B is changed.
We conclude this section with some important properties of the inverse.
Theorem 2.41: Inverses of Transposes and Products
Let A,B,andA i for i = 1,...,k be n × n matrices.
1. If A is an invertible matrix, then (A T ) −1 =(A −1 ) T
2. If A and B are invertible matrices, then AB is invertible and (AB) −1 = B −1 A −1
3. If A 1 ,A 2 ,...,A k are invertible, then the product A 1 A 2 ···A k is invertible, and (A 1 A 2 ···A k ) −1 =
A −1
k
A−1 k−1 ···A−1 2 A−1 1
Consider the following theorem.
Theorem 2.42: Properties of the Inverse
Let A be an n × n matrix and I the usual identity matrix.
1. I is invertible and I −1 = I
2. If A is invertible then so is A −1 ,and(A −1 ) −1 = A
3. If A is invertible then so is A k ,and(A k ) −1 =(A −1 ) k
4. If A is invertible and p is a nonzero real number, then pA is invertible and (pA) −1 = 1 p A−1