### 1 The inverse of a square matrix

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

$\phantom{\rule{2em}{0ex}}k×{k}^{-1}={k}^{-1}×k=1$

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

$\phantom{\rule{2em}{0ex}}AB=BA=I$

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

##### Example 6

Show that the inverse matrix of $A=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right]$ is $B=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -\frac{1}{2}\hfill \\ \hfill 1\hfill & \hfill -\frac{1}{2}\hfill \end{array}\right]$

##### Solution

All we need do is to check that $AB=BA=I$ .

$AB=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right]×\frac{1}{2}\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right]×\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill -1\hfill \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

The reader should check that $BA=I$ also.

We make three important remarks:

$\bullet$ Non-square matrices do not have inverses.

$\bullet$ The inverse of $A$ is usually written ${A}^{-1}$ .

$\bullet$ Not all square matrices have inverses.

Consider $A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$ , and let $B=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]$ be a possible inverse of $A$ .

1. Find $AB$ and $BA$ :

$AB=\left[\begin{array}{cc}\hfill a& \hfill b\\ \hfill 2a& \hfill 2b\end{array}\right]\phantom{\rule{1em}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}BA=\left[\begin{array}{cc}\hfill a+2b\hfill & \hfill 0\hfill \\ \hfill c+2d\hfill & \hfill 0\hfill \end{array}\right]$

2. Equate the elements of $AB$ to those of $I=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ and solve the resulting equations:

$a=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}b=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}2a=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}2b=1$ . Hence $a=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}b=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}a=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}b=\frac{1}{2}$ . 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 .

$\bullet$ 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,

$\phantom{\rule{2em}{0ex}}AB=BA=I$    and    $AC=CA=I$

Consider the two ways of forming the product $CAB$

1. $CAB=C\left(AB\right)=CI=C$
2. $CAB=\left(CA\right)B=IB=B.$

Hence $B=C$ and the inverse is unique .

$\bullet$ There is no such operation as division in matrix algebra.

We do not write $\frac{B}{A}$ but rather

$\phantom{\rule{2em}{0ex}}{A}^{-1}B\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}B{A}^{-1},$

depending on the order required.

$\bullet$ Assuming that the square matrix $A$ has an inverse ${A}^{-1}$ then the solution of

the system of equations $AX=B$ is found by pre-multiplying both sides by ${A}^{-1}$ .