3. Inverse Matrices
For numbers, the multiplicative inverse of a nonzero
number a is the reciprocal 1/a. The most important
property of 1/a is that 1
a
* a = a *
1
a
= 1
4. Inverse Matrices
For numbers, the multiplicative inverse of a nonzero
number a is the reciprocal 1/a. The most important
property of 1/a is that 1
a
* a = a *
1
a
= 1
The most important usage of the inverse 1/a
is to solve the equation ax = b.
5. Inverse Matrices
For numbers, the multiplicative inverse of a nonzero
number a is the reciprocal 1/a. The most important
property of 1/a is that 1
a
* a = a *
1
a
= 1
The most important usage of the inverse 1/a
is to solve the equation ax = b. For example, to solve
2x = 3
6. Inverse Matrices
For numbers, the multiplicative inverse of a nonzero
number a is the reciprocal 1/a. The most important
property of 1/a is that 1
a
* a = a *
1
a
= 1
The most important usage of the inverse 1/a
is to solve the equation ax = b. For example, to solve
2x = Multiply both sides by ½½ * ½ * 3
x = 3/2
7. Inverse Matrices
For numbers, the multiplicative inverse of a nonzero
number a is the reciprocal 1/a. The most important
property of 1/a is that 1
a
* a = a *
1
a
= 1
The most important usage of the inverse 1/a
is to solve the equation ax = b. For example, to solve
2x = Multiply both sides by ½½ *
x = 3/2
Let A be an n x n matrix.
½ * 3
8. Inverse Matrices
For numbers, the multiplicative inverse of a nonzero
number a is the reciprocal 1/a. The most important
property of 1/a is that 1
a
* a = a *
1
a
= 1
The most important usage of the inverse 1/a
is to solve the equation ax = b. For example, to solve
2x = Multiply both sides by ½½ *
x = 3/2
Let A be an n x n matrix.
Since In plays the role of 1 in multiplication,
we say B is the inverse of A if AB = BA = In
and we write B as A-1.
½ * 3
9. Inverse Matrices
For numbers, the multiplicative inverse of a nonzero
number a is the reciprocal 1/a. The most important
property of 1/a is that 1
a
* a = a *
1
a
= 1
The most important usage of the inverse 1/a
is to solve the equation ax = b. For example, to solve
2x = Multiply both sides by ½½ *
x = 3/2
We can use A-1 to solve matrix equations AX = C,
i.e. to find a matrix X where A and C are matrices.
Let A be an n x n matrix.
Since In plays the role of 1 in multiplication,
we say B is the inverse of A if AB = BA = In
and we write B as A-1.
½ * 3
10. 1 0 0 . . 0
0 1 0 . . .
0 0 1 . . .
0 0 . . . 1
. . . . . .
Inverse Matrices
AA-1 = A-1A = = In
The matrices A, A-1 is a pair of inverse matrices if
11. 1 0 0 . . 0
0 1 0 . . .
0 0 1 . . .
0 0 . . . 1
. . . . . .
Inverse Matrices
AA-1 = A-1A = = In
Where as the inverse of nonzero numbers always
exist, this is not the case for matrices.
The matrices A, A-1 is a pair of inverse matrices if
12. 1 0 0 . . 0
0 1 0 . . .
0 0 1 . . .
0 0 . . . 1
. . . . . .
Inverse Matrices
AA-1 = A-1A = = In
Where as the inverse of nonzero numbers always
exist, this is not the case for matrices.
There're two types of matrices with regard to inverses.
The matrices A, A-1 is a pair of inverse matrices if
13. 1 0 0 . . 0
0 1 0 . . .
0 0 1 . . .
0 0 . . . 1
. . . . . .
Inverse Matrices
AA-1 = A-1A = = In
I. The ones that have inverses, these are called
invertible (or nonsingular) matrices
The matrices A, A-1 is a pair of inverse matrices if
Where as the inverse of nonzero numbers always
exist, this is not the case for matrices.
There're two types of matrices with regard to inverses.
14. 1 0 0 . . 0
0 1 0 . . .
0 0 1 . . .
0 0 . . . 1
. . . . . .
Inverse Matrices
AA-1 = A-1A = = In
I. The ones that have inverses, these are called
invertible (or nonsingular) matrices
II. The ones that don‘t have an inverse are called
noninvertible (or singular) matrices
The matrices A, A-1 is a pair of inverse matrices if
Where as the inverse of nonzero numbers always
exist, this is not the case for matrices.
There're two types of matrices with regard to inverses.
15. An example of a noninvertible matrix is
Inverse Matrices
0 -2
0 3
16. Inverse Matrices
because for any matrix ,
0 -2
0 3
a b
c d
=
0 #
0 #
a b
c d
An example of a noninvertible matrix is
0 -2
0 3
17. Inverse Matrices
because for any matrix ,
0 -2
0 3
a b
c d
=
0 #
0 #
=
1 0
0 1
a b
c d
An example of a noninvertible matrix is
0 -2
0 3
18. Inverse Matrices
because for any matrix ,
0 -2
0 3
a b
c d
=
0 #
0 #
=
1 0
0 1
a b
c d
Method for Finding Inverse Matrices:
An example of a noninvertible matrix is
0 -2
0 3
19. Inverse Matrices
To find the inverse, A-1 of a matrix A, extend the matrix
A by the identity matrix I on the right as shown,
because for any matrix ,
0 -2
0 3
a b
c d
=
0 #
0 #
=
1 0
0 1
a b
c d
Method for Finding Inverse Matrices:
A A I
An example of a noninvertible matrix is
0 -2
0 3
20. Inverse Matrices
To find the inverse, A-1 of a matrix A, extend the matrix
A by the identity matrix I on the right as shown,
because for any matrix ,
0 -2
0 3
a b
c d
=
0 #
0 #
=
1 0
0 1
a b
c d
Method for Finding Inverse Matrices:
A A I
Apply row operations to tranform it so the identity
matrix I is on the left side as , then B = A-1.I B
An example of a noninvertible matrix is
0 -2
0 3
21. Example A: Find the inverse matrix A-1 if
Inverse Matrices
a. A =
1 2
1 3
22. 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
23. 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:
24. 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
25. 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
26. 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
27. 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
(-2)R2 Add to R1
28. 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
(-2)R2 Add to R1
29. 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
30. 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
31. 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 .
32. 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 .
35. 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.
36. 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.
37. 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
38. 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
55. Example B:
a. Write the following system as a matrix equation
x + 2y = 4
x + 3y = -1
An Application of Inverse Matrices
56. Example B:
a. Write the following system as a matrix equation
x + 2y = 4
x + 3y = -1
The system may be expressed as
1 2
1 3
x
y =
4
-1
An Application of Inverse Matrices
57. 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
58. 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.
59. To solve the equation AX = B for the matrix X,
we multiply to the left of both sides by A-1 (if it exists)
An Application of Inverse Matrices
60. 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
An Application of Inverse Matrices
61. 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
An Application of Inverse Matrices
62. 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
An Application of Inverse Matrices
63. 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.
An Application of Inverse Matrices
64. 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
65. Multiply A-1 to the left on both sides of the equation
Inverse Matrices
1 2
1 3
x
y =
4
-1
66. 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
67. 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
68. 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.
69. 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.
70. 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
71. 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
