If $A=\left[a_{i j}\right] 2 \times 2$, where $a_{i j}=\left\{\begin{array}{l}i+j, \text { if } i \neq j \\ i^{2}-2 j, \text { if } i=j\end{array}\right.$, then $A^{-1}=$
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, $A^{-1}=\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$.