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

$$\begin{gathered} adj. A=-A \\ \Rightarrow A\cdot adj.A=-A^2=\left|A\right|I=-I \\ \Rightarrow A^2=I \end{gathered}$$

Let  $$\begin{gathered} A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}{a}^2+bc& (a+d)b\\ (a+d)c& d^2+bc\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$
On comparing both sides, we get
    $a^2+bc=1,(a+b)b=0,(a+d)c=0,d^2+bc=1$
Case I: When  $(a+d)\ne 0$
$\Rightarrow b=0=c$   and  $a=1,d=1$  or  $a=-1,d=-1$

$\therefore A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$

But both are rejected as det. $A=-1$ (given)
Case II: When  $(a+d)=0\Rightarrow bc=0$
(i)    If  $a=1,d=-1,\Rightarrow d=-a$
         For  $b=0,c$ can be -1, 0, 1.
         For $b=1,c$  can be 0 only
         For $b=-1,c$  can be 0 only
         So, 5 matrices are possible.
(ii)    If  $a=-1,d=1$
          $\Rightarrow bc=0$    
         For $b=0,c=-1,0,1$  only.
         For  $b=1,c=0$ only.
         For $b=-1,c=0$ only.
         So, 5 matrices are possible.
(iii)   If  $a=0,d=0$
          $\Rightarrow bc=1$    
         $\therefore A=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]orA=\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]$     
         So, 2 matrices are possible
Therefore, total number of matrices is 12.