Find the inverse of each of the following matrices.

Solution:

(i) $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{ll}3 & 1 \\ 1 & 2\end{array}\right|=5, C_{12}=-\left|\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right|=-1$ and $C_{13}=\left|\begin{array}{ll}2 & 3 \\ 3 & 1\end{array}\right|=-7$

$C_{21}=-\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|=-1, C_{22}=\left|\begin{array}{ll}1 & 3 \\ 3 & 2\end{array}\right|=-7$ and $C_{23}=-\left|\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right|=5$

$C_{31}=\left|\begin{array}{ll}2 & 3 \\ 3 & 1\end{array}\right|=-7, C_{32}=-\left|\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right|=5$ and $C_{33}=\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|=-1$

$\operatorname{adj} A=\left[\begin{array}{ccc}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\right]^{T}=\left[\begin{array}{ccc}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\right]$

and $|A|=-18$

$\therefore A^{-1}=-\frac{1}{18}\left[\begin{array}{ccc}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\right]$

(ii) $B=\left[\begin{array}{ccc}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{cc}-1 & -1 \\ 3 & -1\end{array}\right|=4, C_{12}=-\left|\begin{array}{cc}1 & -1 \\ 2 & -1\end{array}\right|=-1$ and $C_{13}=\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right|=5$

$C_{21}=-\left|\begin{array}{cc}2 & 5 \\ 3 & -1\end{array}\right|=17, C_{22}=\left|\begin{array}{cc}1 & 5 \\ 2 & -1\end{array}\right|=-11$ and $C_{23}=-\left|\begin{array}{cc}1 & 2 \\ 2 & 3\end{array}\right|=1$

$C_{31}=\left|\begin{array}{cc}2 & 5 \\ -1 & -1\end{array}\right|=3, C_{32}=-\left|\begin{array}{cc}1 & 5 \\ 1 & -1\end{array}\right|=6$ and $C_{33}=\left|\begin{array}{cc}1 & 2 \\ 1 & -1\end{array}\right|=-3$

$\operatorname{adj} B=\left[\begin{array}{ccc}4 & -1 & 5 \\ 17 & -11 & 1 \\ 3 & 6 & -3\end{array}\right]^{T}=\left[\begin{array}{ccc}4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3\end{array}\right]$

and $|B|=27$

$\therefore B^{-1}=\frac{1}{27}\left[\begin{array}{ccc}4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3\end{array}\right]$

(iii) $C=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right|=3, C_{12}=-\left|\begin{array}{cc}-1 & -1 \\ 1 & 2\end{array}\right|=1$ and $C_{13}=\left|\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right|=-1$

$C_{21}=-\left|\begin{array}{cc}-1 & 1 \\ -1 & 2\end{array}\right|=1, C_{22}=\left|\begin{array}{cc}2 & 1 \\ 1 & 2\end{array}\right|=3$ and $C_{23}=-\left|\begin{array}{cc}2 & -1 \\ 1 & -1\end{array}\right|=1$

$C_{31}=\left|\begin{array}{cc}-1 & 1 \\ 2 & -1\end{array}\right|=-1, C_{32}=-\left|\begin{array}{cc}2 & 1 \\ -1 & -1\end{array}\right|=1$ and $C_{33}=\left|\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right|=3$

$\operatorname{adj} C=\left[\begin{array}{ccc}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]^{T}=\left[\begin{array}{ccc}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]$

and $|C|=4$

$\therefore C^{-1}=\frac{1}{4}\left[\begin{array}{ccc}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]$

(iv) $D=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{ll}1 & 0 \\ 1 & 3\end{array}\right|=3, C_{12}=-\left|\begin{array}{cc}5 & 0 \\ 0 & 3\end{array}\right|=-15$ and $C_{13}=\left|\begin{array}{ll}5 & 1 \\ 0 & 1\end{array}\right|=5$

$C_{21}=-\left|\begin{array}{cc}0 & -1 \\ 1 & 3\end{array}\right|=-1, C_{22}=\left|\begin{array}{cc}2 & -1 \\ 0 & 3\end{array}\right|=6$ and $C_{23}=-\left|\begin{array}{ll}2 & 0 \\ 0 & 1\end{array}\right|=-2$

$C_{31}=\left|\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right|=1, C_{32}=-\left|\begin{array}{cc}2 & -1 \\ 5 & 0\end{array}\right|=-5$ and $C_{33}=\left|\begin{array}{ll}2 & 0 \\ 5 & 1\end{array}\right|=2$

$\operatorname{adj} D=\left[\begin{array}{ccc}3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2\end{array}\right]^{T}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$

and $|D|=1$

$\therefore D^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$

(v) $E=\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{rr}-3 & 4 \\ -3 & 4\end{array}\right|=0, C_{12}=-\left|\begin{array}{cc}4 & 4 \\ 3 & 4\end{array}\right|=-4$ and $C_{13}=\left|\begin{array}{cc}4 & -3 \\ 3 & -3\end{array}\right|=-3$

$C_{21}=-\left|\begin{array}{cc}1 & -1 \\ -3 & 4\end{array}\right|=-1, C_{22}=\left|\begin{array}{cc}0 & -1 \\ 3 & 4\end{array}\right|=3$ and $C_{23}=-\left|\begin{array}{cc}0 & 1 \\ 3 & -3\end{array}\right|=3$

$C_{31}=\left|\begin{array}{cc}1 & -1 \\ -3 & 4\end{array}\right|=1, C_{32}=-\left|\begin{array}{cc}0 & -1 \\ 4 & 4\end{array}\right|=-4$ and $C_{33}=\left|\begin{array}{cc}0 & 1 \\ 4 & -3\end{array}\right|=-4$

$\operatorname{adj} E=\left[\begin{array}{ccc}0 & -4 & -3 \\ -1 & 3 & 3 \\ 1 & -4 & -4\end{array}\right]^{T}=\left[\begin{array}{ccc}0 & -1 & 1 \\ -4 & 3 & -4 \\ -3 & 3 & -4\end{array}\right]$

and $|E|=-1$

$\therefore E^{-1}=-1\left[\begin{array}{ccc}0 & -1 & 1 \\ -4 & 3 & -4 \\ -3 & 3 & -4\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right]$

(vi) $F=\left[\begin{array}{ccc}0 & 0 & -1 \\ 3 & 4 & 5 \\ -2 & -4 & -7\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{cc}4 & 5 \\ -4 & -7\end{array}\right|=-8, C_{12}=-\left|\begin{array}{cc}3 & 5 \\ -2 & -7\end{array}\right|=11$ and $C_{13}=\left|\begin{array}{cc}3 & 4 \\ -2 & -4\end{array}\right|=-4$

$C_{21}=-\left|\begin{array}{cc}0 & -1 \\ -4 & -7\end{array}\right|=4, C_{22}=\left|\begin{array}{cc}0 & -1 \\ -2 & -7\end{array}\right|=-2$ and $C_{23}=-\left|\begin{array}{cc}0 & 0 \\ -2 & -4\end{array}\right|=0$

$C_{31}=\left|\begin{array}{cc}0 & -1 \\ 4 & 5\end{array}\right|=4, C_{32}=-\left|\begin{array}{cc}0 & -1 \\ 3 & 5\end{array}\right|=-3$ and $C_{33}=\left|\begin{array}{ll}0 & 0 \\ 3 & 4\end{array}\right|=0$

$\operatorname{adj} F=\left[\begin{array}{ccc}-8 & 11 & -4 \\ 4 & -2 & 0 \\ 4 & -3 & 0\end{array}\right]^{T}=\left[\begin{array}{ccc}-8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0\end{array}\right]$

and $|F|=4$

$\therefore F^{-1}=\frac{1}{4}\left[\begin{array}{ccc}-8 & 4 & 4 \\ 11 & -2 & -3 \\ -4 & 0 & 0\end{array}\right]$

(vii) $G=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha\end{array}\right|=-1, C_{12}=-\left|\begin{array}{cc}0 & \sin \alpha \\ 0 & -\cos \alpha\end{array}\right|=0$ and $C_{13}=\left|\begin{array}{cc}0 & \cos \alpha \\ 0 & \sin \alpha\end{array}\right|=0$

$C_{21}=-\left|\begin{array}{cc}0 & 0 \\ \sin \alpha & -\cos \alpha\end{array}\right|=0, C_{22}=\left|\begin{array}{cc}1 & 0 \\ 0 & -\cos \alpha\end{array}\right|=-\cos \alpha$ and $C_{23}=-\left|\begin{array}{cc}1 & 0 \\ 0 & \sin \alpha\end{array}\right|=-\sin \alpha$

$C_{31}=\left|\begin{array}{cc}0 & 0 \\ \cos \alpha & \sin \alpha\end{array}\right|=0, C_{32}=-\left|\begin{array}{cc}1 & 0 \\ 0 & \sin \alpha\end{array}\right|=-\sin \alpha$ and $C_{33}=\left|\begin{array}{cc}1 & 0 \\ 0 & \cos \alpha\end{array}\right|=\cos \alpha$

$\operatorname{adj} F=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha\end{array}\right]^{T}=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha\end{array}\right]$

and $|F|=-1$

$\therefore F^{-1}=-1\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right]$