(i) To prove $[\operatorname{Adj} \mathrm{A}]^{-1}=\operatorname{Adj}\left(\mathrm{A}^{-1}\right)$
Now $A=\left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$
$\begin{aligned}
&|\mathrm{A}|=1(15-1)+2(-10-1)+1(-2-3) \\
&=14-22-5=-13 \neq 0 \\
&\therefore \quad \mathrm{A}^{-1} \text { exists } \\
&\begin{array}{l}
\text { Adj } \mathrm{A}=\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -5 & -1
\end{array}\right] \\
\therefore \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\text { Adj A) }
\end{array} \\
&=-\frac{1}{13}\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -5 & -1
\end{array}\right]
\end{aligned}$
Now, let B $=$ Adj A
To find $[\operatorname{Adj} A]^{-1}$, we find $\mathrm{B}^{-1}$
Now, $B=\operatorname{Adj} A=\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -5 & -1\end{array}\right]$
$|\mathrm{B}|=14(-4-9)-11(-11-15)-3(-33+20)$
$=-182+286+65=169$
Adj B $=\left[\begin{array}{ccc}-13 & -26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]$
$\mathrm{B}^{-1}=\frac{1}{\mathrm{~B}}$ Adj B $=\frac{1}{169}\left[\begin{array}{ccc}-13 & -26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]$
Thus (Adj A) ${ }^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$
$=$ L.H.S.
Now A ${ }^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$
$\operatorname{Adj}\left(\mathrm{A}^{-1}\right)=\frac{-1}{169}\left[\begin{array}{ccc}-13 & -26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]$
$=$ R.H.S. $=\frac{1}{-13}\left[\begin{array}{ccc}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]$
Now, let $\mathrm{C}=\mathrm{A}^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$
$=\left[\begin{array}{ccc}-\frac{14}{13} & -\frac{11}{13} & -\frac{5}{13} \\ -\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13}\end{array}\right]$
$\operatorname{Adj}\left(A^{-1}\right)=$ Adj. $C\left[\begin{array}{ccc}-\frac{1}{13} & \frac{2}{13} & -\frac{1}{13} \\ \frac{2}{13} & -\frac{3}{13} & -\frac{1}{13} \\ -\frac{1}{13} & -\frac{1}{13} & -\frac{5}{13}\end{array}\right]$
$=\frac{1}{13}\left[\begin{array}{ccc}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]$
Hence $\left(\right.$ Adj. $\left.\mathrm{A}^{-1}\right)=\left(\right.$ Adj. $\left.\mathrm{A}^{-1}\right)$
(ii) Given $\mathrm{A}=\left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$
and $\mathrm{A}^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$.... (Proved above)
Let $\mathrm{C}=\mathrm{A}^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$
$\begin{aligned}
&=\left[\begin{array}{ccc}
-\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\
-\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{3}{13} & \frac{1}{13}
\end{array}\right] \\
&|\mathrm{C}|=\mathrm{a}_{11} \mathrm{~A}_{11}+\mathrm{a}_{21} \mathrm{~A}_{21}+\mathrm{a}_{31} \mathrm{~A}_{31} \\
&=-\frac{14}{13} \times-\frac{1}{13}+\frac{11}{13} \times \frac{2}{13}+\frac{5}{13} \times \frac{-1}{13} \\
&=\frac{14-22-5}{169}=\frac{-13}{169}=\frac{-1}{13} \neq 0
\end{aligned}$
$\therefore \quad \mathrm{C}^{-1}$ exists.
$\begin{aligned}
&\therefore \quad \mathrm{C}^{-1}=\frac{1}{|\mathrm{C}|} \text { Adj. C } \\
&=\frac{1}{-\frac{1}{13}} \times \frac{1}{13}\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right] \\
&\Rightarrow\left(\mathrm{A}^{-1}\right)^{-1}=\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right]=\left[\begin{array}{ccc}
1 & -2 & 1 \\
-2 & 3 & 1 \\
-1 & 1 & 5
\end{array}\right]
\end{aligned}$
Hence, $\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}$.