The number of matrices $A=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]$, where $a,b,c,\mathrm{d}\in \left\{-1,0,1,2,3,\dots \dots ,10\right\}$, such that $A=A^{-1}$, is ______.

Given, $A=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]$ and $A=A^{-1}$

So, $A^2=A·A^{-1}=I$

$\Rightarrow \left[\begin{array}{ll}a& b\\ c& d\end{array}\right]\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{ll}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]$

On comparing both side we get,

$\therefore a^2+bc=1 \cdots \left(1\right)$

$ab+bd=0 \cdots \left(2\right)$

$ac+cd=0 \cdots \left(3\right)$

$bc+d^2=1 \cdots \left(4\right)$

Now equation $\left(1\right)-\left(4\right)$ gives

$a^2-d^2=0$

$\Rightarrow \left(a+d\right)=0$ or $a-d=0$

Case-I

$a+d=0\Rightarrow \left(a,d\right)=\left(-1,1\right),\left(0,0\right),\left(1,-1\right)$

Assuming case $\left(a\right)$$\Rightarrow \left(a,d\right)=\left(-1,1\right)$

Now from equation $\left(1\right)$

$1+bc=1\Rightarrow bc=0$

When $b=0, c=12$ possibilities

When $c=0, b=12$ possibilities

But $\left(0,0\right)$ is repeated

$\therefore 2\times 12=24$

So, total case will be $24-1$ (repeated) $=23$ pairs.

case $\left(b\right)$ $\Rightarrow \left(a,d\right)=\left(1,-1\right)\Rightarrow bc=0\to 23$ pairs

case $\left(c\right)\Rightarrow \left(a,d\right)=\left(0,0\right)\Rightarrow bc=1$

$\Rightarrow \left(b,c\right)=\left(1,1\right)$ and $\left(-1,-1\right)\to 2$ pairs

Case-II

When $a=d$

from $\left(2\right)$ and $\left(3\right)$

$a\ne 0$ then $b=c=0$

$a^2=1$

$\Rightarrow a=\pm 1=d$

$\left(a,d\right)=\left(1,1\right),\left(-1,-1\right)\to 2$ pairs

$\therefore$  Total $=23+23+2+2$

$=50$ pairs.