Solve this

Given:

$A=\left[a_{i j}\right]_{2 \times 2}$, where $a_{i j}=\left\{\begin{array}{l}i+j, \text { if } i \neq j \\ i^{2}-2 j, \text { if } i=j\end{array}\right.$

$a_{i j}=\left\{\begin{array}{l}i+j, \text { if } i \neq j \\ i^{2}-2 j, \text { if } i=j\end{array}\right.$

$a_{11}=(1)^{2}-2(1)=-1$

$a_{12}=1+2=3$

$a_{21}=2+1=3$

$a_{22}=(2)^{2}-2(2)=0$

Thus,

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

Therefore,

$A^{-1}=\frac{1}{0-9}\left[\begin{array}{cc}0 & -3 \\ -3 & -1\end{array}\right]$

$=-\frac{1}{9}\left[\begin{array}{cc}0 & -3 \\ -3 & -1\end{array}\right]$

$=\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$

Hence, $A^{-1}=\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$.