1 The inverse of a square matrix
We know that any nonzero 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
$\phantom{\rule{2em}{0ex}}k\times {k}^{1}={k}^{1}\times k=1$
We now show that an inverse of a matrix can, in certain circumstances, also be defined.
Given an $n\times n$ square matrix $A,$ then an $n\times n$ 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 $AB=BA=I$ .
$AB=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]\times \frac{1}{2}\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]\times \left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$
The reader should check that $BA=I$ also.
We make three important remarks:
$\bullet $ Nonsquare matrices do not have inverses.
$\bullet $ The inverse of $A$ is usually written ${A}^{1}$ .
$\bullet $ Not all square matrices have inverses.
Task!
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 $A$ .

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

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 nonsingular .
$\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 $CAB$
 $CAB=C\left(AB\right)=CI=C$
 $CAB=\left(CA\right)B=IB=B.$
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 premultiplying both sides by ${A}^{1}$ .
$$\begin{array}{rcll}AX& =& B& \text{}\\ \text{premultiplyingby}\phantom{\rule{1em}{0ex}}{A}^{1}:\phantom{\rule{2em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{A}^{1}\left(AX\right)& =& {A}^{1}B,& \text{}\\ \text{usingassociativity:}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left({A}^{1}A\right)X& =& {A}^{1}B& \text{}\\ \text{using}\phantom{\rule{1em}{0ex}}{A}^{1}A=I:\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}IX& =& {A}^{1}B,& \text{}\\ \text{usingpropertyof}\phantom{\rule{1em}{0ex}}I:\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}X& =& {A}^{1}B\phantom{\rule{2em}{0ex}}\text{whichisthesolutionweseek.}& \text{}\end{array}$$