cofactor of $0$ in $| A |$ is $2 - 3a$. According to value of $A^{-1}$,
$\frac{{2 - 3a}}{{|A|}}$ $=$ $\frac{1}{2}$
==>$\frac{{2 - 3a}}{{2(a - 2)}}$ $=$ $\frac{1}{2}$
==> $2 - 3a = a - 2$
==> $a = 1$
Again $c =$ $\frac{{cofactor\,\,of\,\,a\,\,in\,\,|A|}}{{|A|}}$
$=$ $\frac{{\left| {\,\begin{array}{*{20}{c}}0&2\\1&3\end{array}\,} \right|}}{{2(a - 2)}}$
$=$ $\frac{2}{{2(1 - 2)}}$
$=$ $- 1$
Alternative : $AA^{-1} = I$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$T_p=\left\{A=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1, \ldots ., \mathrm{p}-1\}\right\}$
$1.$ The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and $\operatorname{det}(\mathrm{A})$ divisible by $\mathrm{p}$ is
$(A)$ $(\mathrm{p}-1)^2$ $(B)$ $2(\mathrm{p}-1)$
$(C)$ $(\mathrm{p}-1)^2+1$ $(D)$ $2 \mathrm{p}-1$
$2.$ The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $p$ is [Note: The trace of a matrix is the sum of its diagonal entries.]
$(A)$ $(\mathrm{p}-1)\left(\mathrm{p}^2-\mathrm{p}+1\right)$ $(B)$ $\mathrm{p}^3-(\mathrm{p}-1)^2$
$(C)$ $(\mathrm{p}-1)^2$ $(D)$ $(p-1)\left(p^2-2\right)$
$3.$ The number of $A$ in $T_p$ such that det $(A)$ is not divisible by $p$ is
$(A)$ $2 \mathrm{p}^2$ $(B)$ $p^3-5 p$ $(C)$ $p^3-3 p$ $(D)$ $p^3-p^2$
Give the answer question $1,2$ and $3.$