Let verify that

Solution:

$A=\left[\begin{array}{rrr}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$

$\therefore|A|=1(15-1)+2(-10-1)+1(-2-3)=14-22-5=-13$

Now,

$A_{11}=14, A_{12}=11, A_{13}=-5$

$A_{21}=11, A_{22}=4, A_{23}=-3$

$A_{31}=-5, A_{17}=-3, A_{13}=-1$

$\therefore \operatorname{adj} A=\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)$

$=-\frac{1}{13}\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc}-14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1\end{array}\right]$

(i)

$\begin{aligned}|\operatorname{adj} A| &=14(-4-9)-11(-11-15)-5(-33+20) \\ &=14(-13)-11(-26)-5(-13) \\ &=-182+286+65=169 \end{aligned}$

We have,

$\operatorname{adj}(\operatorname{adj} A)=\left[\begin{array}{ccc}-13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]$

$\therefore[\operatorname{adj} A]^{-1}=\frac{1}{|\operatorname{adj} A|}(\operatorname{adj}(\operatorname{adj} A))$

$=\frac{1}{169}\left[\begin{array}{lll}-13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]$

$=\frac{1}{13}\left[\begin{array}{lll}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]$

Now, $A^{-1}=\frac{1}{13}\left[\begin{array}{ccc}-14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\ -\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13}\end{array}\right]$

$\therefore \operatorname{adj}\left(A^{-1}\right)=\left[\begin{array}{ccc}-\frac{4}{169}-\frac{9}{169} & -\left(-\frac{11}{169}-\frac{15}{169}\right) & -\frac{33}{169}+\frac{20}{169} \\ -\left(-\frac{11}{169}-\frac{15}{169}\right) & -\frac{14}{169}-\frac{25}{169} & -\left(-\frac{42}{169}+\frac{55}{169}\right) \\ -\frac{33}{169}+\frac{20}{169} & -\left(-\frac{42}{169}+\frac{55}{169}\right) & \frac{56}{169}-\frac{121}{169}\end{array}\right]$

$=\frac{1}{169}\left[\begin{array}{lll}-13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]=\frac{1}{13}\left[\begin{array}{lll}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]$

Hence, $[\operatorname{adj} A]^{-1}=\operatorname{adj}\left(A^{-1}\right)$.

(ii)

We have shown that:

$A^{-1}=\frac{1}{13}\left[\begin{array}{lll}-14 & -11 & 5 \\ -11 & -4 & 3 \\ 5 & 3 & 1\end{array}\right]$

 

And, $\operatorname{adj} A^{-1}=\frac{1}{13}\left[\begin{array}{lll}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]$

Now,

$\left|A^{-1}\right|=\left(\frac{1}{13}\right)^{3}[-14 \times(-13)+11 \times(-26)+5 \times(-13)]=\left(\frac{1}{13}\right)^{3} \times(-169)=-\frac{1}{13}$

$\therefore\left(A^{-1}\right)^{-1}=\frac{\text { adj } A^{-1}}{\left|A^{-1}\right|}=\frac{1}{\left(-\frac{1}{13}\right)} \times \frac{1}{13}\left[\begin{array}{lll}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]=\left[\begin{array}{lll}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]=A$

$\therefore\left(A^{-1}\right)^{-1}=A$