Matrix Inverses

1

1Monday, October 7, 13

Gaussian Elimination

2

5x1 � x2 + 2x3 = 7

�2x1 + 6x2 + 9x3 = 0

�7x1 + 5x2 � 3x3 = 5

Gaussian Elimination is an algorithm for linear systems like the following

In this set of slides, we want to understand how the matrix inverse can be used express the solution in theoretical terms.

2Monday, October 7, 13

Using GE to solve a system2

45 �1 2 7

�2 6 9 0�7 5 �3 5

3

5Apply elementary row operations to the augmented matrix

2

64

5 �1 2 7

0 285

495

145

0 185 � 1

5745

3

75 (eqn 2)�✓�25

◆(eqn 1)

(eqn 3)�✓�75

◆(eqn 1)

2

64

5 �1 2 7

0 285

495

145

0 0 � 6510

655

3

75 (eqn 3)�

✓9

14

◆(eqn 2)

PivotsMultipliers

3

Use back substitution : x1 = 3, x2 = 4, x3 = �2

3Monday, October 7, 13

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!4

4Monday, October 7, 13

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.

55Monday, October 7, 13

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.

66Monday, October 7, 13

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 :

77Monday, October 7, 13

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.

88Monday, October 7, 13

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 99Monday, October 7, 13

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

1010Monday, October 7, 13

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)

1111Monday, October 7, 13

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

1212Monday, October 7, 13

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.

1313Monday, October 7, 13

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.

1414Monday, October 7, 13