inverse matrices

1. Example A: Find the inverse matrix A-1 if
Inverse Matrices
a. A =
1 2
1 3
Extend
1 2
1 3
to
1 2 1 0
1 3 0 1
Apply row operations to tranform it so the identity
matrix I is on the left side:
1 2 1 0
1 3 0 1
(-1)R1 Add to R2
-1 -2 -1 0
1 2 1 0
0 1 -1 1
0 -2 2 -2
1 0 3 -2
0 1 -1 1
(-2)R2 Add to R1
=
identity matrix
Hence A-1 =
3 -2
-1 1 .
One checks
easily that AA-1 = A-1A =
1 0
0 1 .

2. Inverse Matrices
b. A =
0 2
0 3
Extend
0 2
0 3
to
0 2 1 0
0 3 0 1
Since the first column consists of 0's only, there is
no way we may introduce a 1 at the a11 position by
row operations. Hence A-1 doesn't exist.
c. A =
0 1 -1
1 -1 2
1 1 0
Extend
0 1 -1
1 -1 2
1 1 0
0 1 -1 1 0 0
1 -1 2 0 1 0
1 1 0 0 0 1

3. Inverse Matrices
Apply row operations:
0 1 1 1 0 0
1 1 2 0 1 0
1 1 0 0 0 1
(-1)R2 add
to R1
0 1 1 1 0 0
1 1 2 0 1 0
1 1 0 0 0 1
R1 R2
-1 -1 -2 0 -1 0
0 1 1 1 0 0
1 1 2 0 1 0
0 0 -2 0 -1 1
0 -1 -1 -1 0 0
0 1 -1 1 0 0
1 0 1 -1 1 0
0 0 -2 0 -1 1
(-1)R1 add
to R3
0 1 0 1 -1/2 1/2
1 0 0 -1 1/2 1/2
0 1 -1 1 0 0
1 0 1 -1 1 0
0 0 1 0 1/2 -1/2
0 0 -1 0 -1/2 1/2
(-1)R3 + R1
(-1/2)R3
R3 + R2
0 0 1 0 1/2 -1/2
=
A-1We check that A-1*A = A*A-1 = I.
Let’s use an inverse matrix to solve a matrix equation.

4. Example B:
a. Write the following system as a matrix equation
If we let A =
1 2
1 3
x + 2y = 4
x + 3y = -1
, X =
x
y
,
The system may be expressed as
1 2
1 3
x
y =
4
-1
then the matrix equation is simply AX = B.
and B =
4
-1 .
An Application of Inverse Matrices
To solve for the matrix X, we use the following
inverse matrix method.

5. To solve the equation AX = B for the matrix X,
we multiply to the left of both sides by A-1 (if it exists)
AX = B
A-1AX = A-1B
IX = A-1B
or that X = A-1B
b. Solve the matrix equation AX = B from part a.
by the inverse matrix method.
from the previous example we've that
Given that A =
1 2
1 3
3 -2
-1 1 .
A-1 =
An Application of Inverse Matrices

6. Multiply A-1 to the left on both sides of the equation
Inverse Matrices
1 2
1 3
x
y =
4
-1
we get
3 -2
-1 1
1 2
1 3
x
y =
3 -2
-1 1
4
-1
0 1
1 0 x
y =
14
-5
Hence
x
y =
14
-5 or that x = 14, and y = -5.
With inverse matrices, we may do matrix algebra
that is similar to the “traditional” algebra.

7. Inverse Matrices
Exercise A. Find the inverse matrix of each of the following
matrices, if it exits
1 0
0 1
1) 2) 3) 4)1 5
0 1
a 5
0 𝑏
0 𝑏
𝑎 5
3 5
–2 1
5) 6) 7) 8)
1 –2
7 –1
𝑎 𝑏
𝑏 𝑎
𝑎 𝑏
𝑎 𝑏
9) 10) 11) 12)
1 0 0
0 1 0
0 0 1
1 2 3
0 1 4
0 0 1
a 0 0
0 b 0
0 0 c
a 2 3
0 b 4
0 0 c
13) 14)
1 0 2
0 1 0
−1 0 1
1 2 3
0 1 4
−1 2 1

8. 2. Verify the following inverse formula for 2x2 matrices.
𝑎 𝑏
𝑐 𝑑
–1
= 1
ad – bc
𝑑 –𝑏
–𝑐 𝑎
(Hint: you don’t have to compute the inverse, just check that
their product is I so it must be the one and only inverse.)
3. Verify the that (AB) –1 = B –1 A–1.
So if A and B are invertible then AB is invertible.
Ex B. A matrix that has an inverse is said to be invertible.
1. (A–1 is unique)
Verify that if A is invertible, then there is only one A–1.
This means to show that if there are matrices B and C
that invert A so that AB = BA = I and AC = CA = I, then B = C.
4. Verify that if AB is invertible, then A and B must be invertible.
(Hint: Let C is (AB) –1, then A–1 or B –1 can be “solved”.)
So if A or B is not invertible then AB is not invertible.
Inverse Matrices

9. (Answers to the odd problems) Exercise A.
𝐴−1 =
1 0
0 1
1)
𝐴−1 =
1/𝑎 −5/𝑎𝑏
0 1/𝑏
𝐴−1 =
1
13
1 −5
2 3
5)
𝐴−1
=
1
𝑎2 − 𝑏2
𝑎 −𝑏
−𝑏 𝑎
9)
𝐴−1 =
1 0 0
0 1 0
0 0 1
𝐴−1 =
1/a 0 0
0 1/b 0
0 0 1/c
13)
𝐴−1 =
1
3
1 0 −2
0 3 0
1 0 1
3)
7)
11)
Inverse Matrices