# Find the inverse of each of the following matrices:

Question:

Find the inverse of each of the following matrices:

(i) $\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

(ii) $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

(iii) $\left[\begin{array}{cc}a & b \\ c & \frac{1+b c}{a}\end{array}\right]$

(iv) $\left[\begin{array}{cc}2 & 5 \\ -3 & 1\end{array}\right]$

Solution:

(i) $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

$|A|=\cos ^{2} \theta+\sin ^{2} \theta=1 \neq 0$

$A$ is a singular matrix; therefore, it is invertible.

Let $C_{i j}$ be a cofactor of $a_{i j}$ in $A$.

Now,

$C_{11}=\cos \theta$

$C_{12}=\sin \theta$

$C_{21}=-\sin \theta$

$C_{22}=\cos \theta$

$\operatorname{adj} A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]^{T}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$

(ii) $B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

$|B|=0-1=-1 \neq 0$

$B$ is a singular matrix; therefore, it is invertible.

Let $C_{i j}$ be a cofactor of $b_{i j}$ in $B$.

Now,

$C_{11}=0$

$C_{12}=-1$

$C_{21}=-1$

$C_{22}=0$

$\operatorname{adj} B=\left[\begin{array}{cc}0 & -1 \\ -1 & 0\end{array}\right]^{T}=\left[\begin{array}{cc}0 & -1 \\ -1 & 0\end{array}\right]$

$\therefore B^{-1}=\frac{1}{|B|} \operatorname{adj} B=\frac{1}{-1}\left[\begin{array}{cc}0 & -1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$

(iii) $C=\left[\begin{array}{cc}a & b \\ c & \frac{1+b c}{a}\end{array}\right]$

$|C|=1+b c-b c=1 \neq 0$

$C$ is a singular matrix; therefore, it is invertible.

Let $C_{i j}$ be a cofactor of $c_{i j}$ in $C$.

Now,

$C_{11}=\frac{1+b c}{a}$

$C_{12}=-c$

$C_{21}=-b$

$C_{22}=a$

$\operatorname{adj} C=\left[\begin{array}{cc}\frac{1+b c}{a} & -c \\ -b & a\end{array}\right]^{T}=\left[\begin{array}{cc}\frac{1+b c}{a} & -b \\ -c & a\end{array}\right]$

$\therefore C^{-1}=\frac{1}{|C|} \operatorname{adj} C=\left[\begin{array}{cc}\frac{1+b c}{a} & -b \\ -c & a\end{array}\right]$

$(i v) D=\left[\begin{array}{cc}2 & 5 \\ -3 & 1\end{array}\right]$

$|D|=2+15=17 \neq 0$

$D$ is a singular matrix; therefore, it is invertible.

Let $C_{i j}$ be a cofactor of $d_{i j}$ in $D .$

Now,

$C_{11}=1$

$C_{12}=3$

$C_{21}=-5$

$C_{22}=2$

$\operatorname{adj} D=\left[\begin{array}{cc}1 & 3 \\ -5 & 2\end{array}\right]^{T}=\left[\begin{array}{cc}1 & -5 \\ 3 & 2\end{array}\right]$

$\therefore D^{-1}=\frac{1}{|D|} \operatorname{adj} D=\frac{1}{17}\left[\begin{array}{cc}1 & -5 \\ 3 & 2\end{array}\right]$