Question

If $A={\left[a_{ij}\right]}_{2\times 2}$ where $a_{15}=\{\begin{array}{c}i+j\\ i^2-2j\end{array}\begin{array}{c}i\ne j\\ i=j\end{array}$ then $A^{-1}=$

• A. $\frac{1}{9}\left[\begin{array}{cc}0& 3\\ 3& 1\end{array}\right]$
• B. $\frac{1}{9}\left[\begin{array}{cc}0& -3\\ 3& -1\end{array}\right]$
• C. $\frac{1}{9}\left[\begin{array}{cc}0& -3\\ 3& 1\end{array}\right]$
• D. $\frac{1}{3}\left[\begin{array}{cc}0& 3\\ 3& 1\end{array}\right]$

Answer 1

Answer: (A)

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

$A={\left[a_{ij}\right]}_{2\times 2}$

$\Rightarrow a_{15}=\{i+ji\ne j{i}^2-2ji=j\}$

$\Rightarrow a_{11}=1-2=-1,a_{12}=1+2=3$

$\Rightarrow a_{21}=2+1=3,a_{22}=4-4=0$

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

$\Rightarrow A^{-1}=\frac{1}{\left|A\right|}adjA$

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

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

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

Hence, the answer is $\frac{1}{9}\left[\begin{array}{cc}0& 3\\ 3& 1\end{array}\right].$