### 2 The inverse of a 2 $×$ 2 matrix

In this subsection we show how the inverse of a $2×2$ matrix can be obtained (if it exists).

Form the matrix products $AB$ and $BA$ where

$A=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}B=\left[\begin{array}{cc}\hfill d& \hfill -b\\ \hfill -c& \hfill a\end{array}\right]$

$\phantom{\rule{2em}{0ex}}AB=\left[\begin{array}{cc}\hfill ad-bc\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill ad-bc\hfill \end{array}\right]=\left(ad-bc\right)\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left(ad-bc\right)I$

$\phantom{\rule{2em}{0ex}}BA=\left[\begin{array}{cc}\hfill ad-bc\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill ad-bc\hfill \end{array}\right]=\left(ad-bc\right)I$

You will see that had we chosen $C=\frac{1}{ad-bc}$ $\left[\begin{array}{cc}\hfill d& \hfill -b\\ \hfill -c& \hfill a\end{array}\right]$ instead of $B$ then both products $AC$ and $CA$ will be equal to $I$ . This requires $ad-bc\ne 0$ . Hence this matrix $C$ is the inverse of $A$ . However, note, that if $ad-bc=0$ then $A$ has no inverse . (Note that for the matrix $A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right],$ which occurred in the last task, $ad-bc=1×0-0×2=0$ confirming, as we found, that $A$ has no inverse.)

##### Key Point 8

The Inverse of a 2 $×$ 2 Matrix

If $ad-bc\ne 0$ then the $2×2$ matrix $A=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]$ has a (unique) inverse given by

${A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}\hfill d& \hfill -b\\ \hfill -c& \hfill a\end{array}\right]$

Note that $ad-bc=\left|A\right|$ , the determinant of the matrix $A$ .

In words: To find the inverse of a $2×2$ matrix $A$ we interchange the diagonal elements, change the sign of the other two elements, and then divide by the determinant of $A$ .

Which of the following matrices has an inverse?

$\phantom{\rule{2em}{0ex}}A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right],\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}B=\left[\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill -1& \hfill 1\end{array}\right],\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}C=\left[\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill -2& \hfill 2\end{array}\right],\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}D=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$|A|=1×3-0×2=3$ ; $|B|=1+1=2;$ $|C|=2-2=0;$ $|D|=1-0=1.$

Therefore, $A,\phantom{\rule{1em}{0ex}}B$ and $D$ each has an inverse. $C$ does not because it has a zero determinant.

Find the inverses of the matrices $A,\phantom{\rule{1em}{0ex}}B$ and $D$ in the previous Task.

Use Key Point 8:

${A}^{-1}=\frac{1}{3}\left[\begin{array}{cc}\hfill 3& \hfill 0\\ \hfill -2& \hfill 1\end{array}\right],\phantom{\rule{1em}{0ex}}{B}^{-1}=\frac{1}{2}\left[\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right],\phantom{\rule{1em}{0ex}}{D}^{-1}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=D$

It can be shown that the matrix $A=\left[\begin{array}{cc}\hfill cos\theta & \hfill sin\theta \\ \hfill -sin\theta & \hfill cos\theta \end{array}\right]$ represents an anti-clockwise rotation through an angle $\theta$ in an $xy$ -plane about the origin. The matrix $B$ represents a rotation clockwise through an angle $\theta$ . It is given therefore by

$\phantom{\rule{2em}{0ex}}B=\left[\begin{array}{cc}\hfill cos\left(-\theta \right)& \hfill sin\left(-\theta \right)\\ \hfill -sin\left(-\theta \right)& \hfill cos\left(-\theta \right)\end{array}\right]=\left[\begin{array}{cc}\hfill cos\theta & \hfill -sin\theta \\ \hfill sin\theta & \hfill cos\theta \end{array}\right]$

Form the products $AB$ and $BA$ for these ‘rotation matrices’. Confirm that $B$ is the inverse matrix of $A$ .

$\begin{array}{rcll}AB& =& \left[\begin{array}{cc}\hfill cos\theta & \hfill sin\theta \\ \hfill -sin\theta & \hfill cos\theta \end{array}\right]\left[\begin{array}{cc}\hfill cos\theta & \hfill -sin\theta \\ \hfill sin\theta & \hfill cos\theta \end{array}\right]& \text{}\\ & =& \left[\begin{array}{cc}\hfill {cos}^{2}\theta +{sin}^{2}\theta \hfill & \hfill -cos\theta sin\theta +sin\theta cos\theta \hfill \\ \hfill -sin\theta cos\theta +cos\theta sin\theta \hfill & \hfill {sin}^{2}\theta +{cos}^{2}\theta \hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=I& \text{}\end{array}$

Similarly, $BA=I$

Effectively: a rotation through an angle $\theta$ followed by a rotation through angle $-\theta$ is equivalent to zero rotation.