Matrix Inverses

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Matrix inverse

3x = 7

We can solve the linear system

by multiplying both sides by the “inverse” of 3 :

(3�1)3x = (3�1)7

x =7

3

If a linear system is non-singular or invertible, we can write the solution in the same manner

Ax = b

x = A�1b

where A�1is the inverse of A.

solution is unique!

solution is unique!���2

The identity matrix

2

44 �3 21 0 �52 7 �9

3

5

2

41 0 00 1 00 0 1

3

5 =

2

44 �3 21 0 �52 7 �9

3

5

The identity matrix is the “multiplicative identity”, or the “1” for square matrices.

2

41 0 00 1 00 0 1

3

5

2

44 �3 21 0 �52 7 �9

3

5 =

2

44 �3 21 0 �52 7 �9

3

5Ex.

It has the property that for any n⇥ n matrix A,

AI = IA = A

An n ⇥ n identity matrix I is a diagonal matrix with

all 1s on the diagonal.

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Matrix inverse

Just as (3

�1)(3) = (3)(3

�1) = 1, a matrix A and its

inverse satisfy

A�1A = AA�1= I

where I is the identity matrix.

2

64

0 1 2

1 0 3

4 �3 8

3

75

2

64

� 92 7 � 3

2

�2 4 �132 �2 � 1

2

3

75 =

A A�1

2

64

1 0 0

0 1 0

0 0 1

3

75

I

2

64

� 92 7 � 3

2

�2 4 �132 �2 � 1

2

3

75

2

64

0 1 2

1 0 3

4 �3 8

3

75 =

AA�1

2

64

1 0 0

0 1 0

0 0 1

3

75

I

The identity times a vector also returns that vector.

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Matrix inverse

Multiply both sides by A�1to get

A�1Ax = A�1b

Ix = A�1b

Since I times any vector or matrix is always that vector

or matrix, we have

x = A�1b

The solution is unique

If A is invertible, we could solve Ax = b as follows :

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Example2

64

0 1 2

1 0 3

4 �3 8

3

75

2

64x1

x2

x3

3

75 =

2

64�1

2

11

3

75

Multiply both sides by A�1to get

Check!

2

64x1

x2

x3

3

75 =

2

64

� 92 7 � 3

2

�2 4 �132 �2 � 1

2

3

75

2

64�1

2

11

3

75 =

2

642

�1

0

3

75

A x b

A�1x b

This is the only solution to this system.

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Two special cases

The inverse of a 2 ⇥ 2 matrix can easily be written

down.

a bc d

��1

=

1

ad� bc

d �b

�c a

�

The inverse of a diagonal matrix is given by

2

6664

d1 0

d2.

.

.

0 dn

3

7775

�1

=

2

6664

1d1

0

1d2

.

.

.

0

1dn

3

7775

the “determinant”

All off-diagonal entries are zero ���7

Gauss-Jordan Method

2

40 1 2 1 0 01 0 3 0 1 04 �3 8 0 0 1

3

5

(eqn 2)

(eqn 1)

2

41 0 3 0 1 00 1 2 1 0 04 �3 8 0 0 1

3

5

2

41 0 3 0 1 00 1 2 1 0 00 �3 �4 0 �4 1

3

5 (eqn 3)� (4)(eqn 1)

2

41 0 3 0 1 00 1 2 1 0 00 0 2 3 �4 1

3

5 (eqn 3)� (�3)(eqn 2)

A I

If we were solving a linear system, we could stop here at the upper triangular form. But to get the inverse, we continue. upper triangular

The Gauss-Jordon method gives us a way to compute the matrix inverse.

Carry out row operations on A and I simultaneously. The first step is a row

exchange

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Gauss-Jordan Method

Divide the last row by 2 2

41 0 3 0 1 00 1 2 1 0 00 0 1 3

2 �2 12

3

5

Reverse the elimination process until A is reduced to the identity.

2

64

1 0 0 � 92 7 � 3

2

0 1 0 �2 4 �1

0 0 1 32 �2 1

2

3

75 (eqn 2)� (2)(eqn 3)

(eqn 1)� (3)(eqn 3)

I A�1

The augmented matrix [A I] is row-reduced to [I A�1].

The Gauss-Jordan Method is a method for finding the inverse of a matrix.

get 1s on the diagonal ✓1

2

◆(eqn 3)

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Example

Find the inverse of A by two di↵erent methods. Verify

the inverse you found.

A =

1 3

2 7

�

Use the formula for the inverse of a 2x2 matrix

A�1 =1

(1)(7)� (3)(2)

7 �3

�2 1

�=

7 �3

�2 1

�

1 3 1 02 7 0 1

�!

1 3 1 00 1 �2 1

�!

1 0 7 �30 1 �2 1

�Use the Gauss-Jordan Method

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The inverse of sums and products

Suppose both A and B are invertible. What can we say about the inverse of(A+B)? Or the inverse of AB?

We can’t say anything about the inverse of A + B. Suppose A is invertible. Let

B = �A. Then A+B = 0, which is not invertible.

If A and B are invertible, the inverse of the product AB is given by (AB)

�1=

B�1A�1.

(AB)(AB)

�1= ABB�1A�1

= AIA�1= I

We don’t need to check the other side, since the inverse is unique and so is both

a left and right inverse.

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Some facts about inverses

The inverse exists if and only if elimination produces n (non-zero) pivots.

A matrix cannot have two di↵erent inverses. If you find one inverse, you have

found both the right and left inverse.

A matrix A commutes with its inverse : AA�1= A�1A = I

If A is invertible, then Ax = 0 can only have the zero solution x = A�10 = 0.

Suppose there is a non-zero vector x such that Ax = 0. Then A cannot have an

inverse. Suppose it did have an inverse B such that AB = BA = I. Then

x = B0 = 0

But we said that x is nonzero. Therefore, the inverse B cannot exist. Matrices

which don’t have inverses are called singular or non-invertible. Such matrices are

like the multiplicative ”0” of matrices. (Think of the scalar system 0x = b. We

cannot divide by 0 to solve for x.)

Only square matrices have inverses.

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