Find the inverse of each of the following matrices

Solution:

(i) $A=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{ll}4 & 3 \\ 3 & 4\end{array}\right|=7, C_{12}=-\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array}\right|=-1$ and $C_{13}=\left|\begin{array}{cc}1 & 4 \\ 1 & 3\end{array}\right|=-1$

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

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

$\operatorname{adj} A=\left[\begin{array}{ccc}7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1\end{array}\right]^{T}=\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$

and $|A|=1$

$A^{-1}=\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$

Now, $A^{-1} A=\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I_{3}$

(ii) $B=\left[\begin{array}{lll}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{ll}4 & 1 \\ 7 & 2\end{array}\right|=1, C_{12}=-\left|\begin{array}{ll}3 & 1 \\ 3 & 2\end{array}\right|=-3$ and $C_{13}=\left|\begin{array}{ll}3 & 4 \\ 3 & 7\end{array}\right|=9$

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

$C_{31}=\left|\begin{array}{ll}3 & 1 \\ 4 & 1\end{array}\right|=-1, C_{32}=-\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|=1$ and $C_{33}=\left|\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right|=-1$

$\operatorname{adj} B=\left[\begin{array}{ccc}1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1\end{array}\right]^{T}=\left[\begin{array}{ccc}1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1\end{array}\right]$

and $|B|=2$

$B^{-1}=\frac{1}{2}\left[\begin{array}{ccc}1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1\end{array}\right]$

Now, $B^{-1} B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I_{3}$