Question

Let A =$\left[a_{ij}\right]$ be a matrix of order 2 where adjA = - A and $a_{ij}\in \{-2,-1,0,1,2\}$ If det (A) = - 1, then number of such matrices are

Answer 1

$A^{-1}=\frac{adjA}{|A|}\Rightarrow adjA=-A^{-1}$
$A=A^{-1}$
$A^{-1}=\frac{\left[\begin{array}{cc}{a}_{22}& -a_{12}\\ -a_{21}& a_{11}\end{array}\right]}{|A|}\Rightarrow A^{-1}=\left[\begin{array}{cc}-a_{22}& a_{12}\\ a_{21}& -a_{11}\end{array}\right]$
$A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$
$A+A^{-1}$
$\Rightarrow a_{11}+a_{22}=0,|A|=a_{11}{a}_{22}-a_{12}{a}_{21}=-1$
$-a_{11}^2-a_{12}{a}_{21}=-1$
$a_{11}^2+a_{12}{a}_{21}=1$
case 1:
$(a_{11}{)}^2$ =1
$a_{11}=1$, or $-1\to 2$ ways
$\begin{array}{c}\left(a\right)a_{12}=0,a_{21}=\underset{}{-2,-1,0,1,2}\\ \left(b\right)a_{12}=\underset{}{1,-1,2,-2},a_{21}=0\end{array}\}9ways$
Total $=2\times 9=18$ ways
case 2:
$a_{11}=0,\begin{array}{c}{a}_{12}=a_{21}=1\\ a_{12}=a_{21}=-1\end{array}\}2$ ways
Total $\to 18+2=20$ ways.