1 The inverse of a square matrix

We know that any non-zero number k has an inverse; for example 2 has an inverse 1 2 or 2 1 . The inverse of the number k is usually written 1 k or, more formally, by k 1 . This numerical inverse has the property that

k × k 1 = k 1 × k = 1

We now show that an inverse of a matrix can, in certain circumstances, also be defined.

Given an n × n square matrix A , then an n × n square matrix B is said to be the inverse matrix of A if

A B = B A = I

where I is, as usual, the identity matrix (or unit matrix) of the appropriate size.

Example 6

Show that the inverse matrix of A = 1 1 2 0 is B = 0 1 2 1 1 2

Solution

All we need do is to check that A B = B A = I .

A B = 1 1 2 0 × 1 2 0 1 2 1 = 1 2 1 1 2 0 × 0 1 2 1 = 1 2 2 0 0 2 = 1 0 0 1

The reader should check that B A = I also.

We make three important remarks:

Non-square matrices do not have inverses.

The inverse of A is usually written A 1 .

Not all square matrices have inverses.

Task!

Consider A = 1 0 2 0 , and let B = a b c d be a possible inverse of A .

  1. Find A B and B A :

    A B = a b 2 a 2 b , B A = a + 2 b 0 c + 2 d 0

  2. Equate the elements of A B to those of I = 1 0 0 1 and solve the resulting equations:

    a = 1 , b = 0 , 2 a = 0 , 2 b = 1 . Hence a = 1 , b = 0 , a = 0 , b = 1 2 . This is not possible!

Hence, we have a contradiction. The matrix A therefore has no inverse and is said to be a singular matrix . A matrix which has an inverse is said to be non-singular .

If a matrix has an inverse then that inverse is unique.

Suppose B and C are both inverses of A . Then, by definition of the inverse,

A B = B A = I    and    A C = C A = I

Consider the two ways of forming the product C A B

  1. C A B = C ( A B ) = C I = C
  2. C A B = ( C A ) B = I B = B .

Hence B = C and the inverse is unique .

There is no such operation as division in matrix algebra.

We do not write B A but rather

A 1 B or B A 1 ,

depending on the order required.

Assuming that the square matrix A has an inverse A 1 then the solution of

the system of equations A X = B is found by pre-multiplying both sides by A 1 .

A X = B pre-multiplying by A 1 : A 1 ( A X ) = A 1 B , using associativity: ( A 1 A ) X = A 1 B using A 1 A = I : I X = A 1 B , using property of I : X = A 1 B which is the solution we seek.