The number of matrices $\mathrm{A}=\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]$, where $\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \{-1,0,1,2,3,\dots \dots ,10\}$, such that $\mathrm{A}={\mathrm{A}}^{-1}$, is______.

$\mathrm{A}=\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]$

Given $\mathrm{A}={\mathrm{A}}^{-1}$

$\begin{array}{l}\therefore {\mathrm{A}}^2=\mathrm{A}\cdot {\mathrm{A}}^{-1}=\mathrm{I}\\ \left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]\left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ \Rightarrow \left[\begin{array}{ll}{\mathrm{a}}^2+\mathrm{bc}& \mathrm{ab}+\mathrm{bd}\\ \mathrm{ac}+\mathrm{cd}& \mathrm{bc}+{\mathrm{d}}^2\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ \therefore {\mathrm{a}}^2+\mathrm{bc}=1 ...\left(1\right)\\ \mathrm{ab}+\mathrm{bd}=0 ...\left(2\right)\\ \mathrm{ac}+\mathrm{cd}=0 ...\left(3\right)\\ \mathrm{bc}+{\mathrm{d}}^2=1 ...\left(4\right)\end{array}$

(1) – (4) gives

$\begin{array}{l}{\mathrm{a}}^2-{\mathrm{d}}^2=0\\ \Rightarrow (\mathrm{a}+\mathrm{d})=0\text{ or }\mathrm{a}-\mathrm{d}=0\end{array}$

Case – I

$$\begin{gathered} \mathrm{a}+\mathrm{d}=0 \Rightarrow (\mathrm{a}, \mathrm{d})=(-1, 1), (0, 0), (1, -1) \\ \left(\mathrm{a}\right) (\mathrm{a}, \mathrm{d})=(-1, 1) \\ \therefore \mathrm{from} \mathrm{equation} \left(1\right) \\ 1+\mathrm{bc}=1 \Rightarrow \mathrm{bc}=0 \\ \mathrm{b} = 0 \mathrm{C} = 12 \mathrm{possibilities} \\ \mathrm{c} = 0 \mathrm{b} = 12 \mathrm{possibilities} \\ \mathrm{but} (0, 0) \mathrm{is} \mathrm{repeated} \\ \therefore 2\times 12=24 \\ \Rightarrow 24 – 1 \left(\mathrm{repeated}\right) = 23 \mathrm{pairs} \\ \left(\mathrm{b}\right) (\mathrm{a},\mathrm{d})=(1, -1) \Rightarrow \mathrm{bc}=0\to 23 \\ \left(\mathrm{c}\right) (\mathrm{a},\mathrm{d})=(0, 0) \Rightarrow \mathrm{bc}=1 \\ \Rightarrow (\mathrm{b}, \mathrm{c})=(1, 1) \mathrm{\&} (-1, -1), 2\text{ pairs } \end{gathered}$$

Case – II

$\mathrm{a}=\mathrm{d}$

from (2) and (3)

$\begin{array}{l}\mathrm{a}\ne 0\text{ then }\mathrm{b}=\mathrm{c}=0\\ {\mathrm{a}}^2=1\\ \mathrm{a}=\pm 1=\mathrm{d}\\ (\mathrm{a}, \mathrm{d})=(1, 1), (-1, -1)\to 2\text{ pairs }\\ \therefore \text{ Total }=23+23+2+2\\ =50\text{ pairs }\end{array}$