The number of matrices A=begin{bmatrix} a & b | vedclass

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(D) Given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $A = A^{-1}$.This implies $A^2 = I$,so $\begin{bmatrix} a & b \ c & d \end{bmatrix}

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$50$

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(D) Given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $A = A^{-1}$.This implies $A^2 = I$,so $\begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.This gives the system of equations:$1) a^2 + bc = 1$$2) b(a + d) = 0$$3) c(a + d) = 0$$4) bc + d^2 = 1$From $(1)$ and $(4)$,$a^2 = d^2$,so $a = d$ or $a = -d$.Case $I$: $a = -d$. Then $b(a - a) = 0$ is always true. We need $a^2 + bc = 1$.If $a = 0$,then $d = 0$ and $bc = 1$. Since $b, c \in \{-1, 0, 1, \ldots, 10\}$,$bc = 1$ implies $(b, c) = (1, 1)$ or $(-1, -1)$. ($2$ pairs).If $a = 1$,then $d = -1$ and $1 + bc = 1 \Rightarrow bc = 0$. This means $b=0$ ($12$ values for $c$) or $c=0$ ($12$ values for $b$). Excluding $(0,0)$ which is counted twice,we have $12 + 12 - 1 = 23$ pairs.If $a = -1$,then $d = 1$ and $1 + bc = 1 \Rightarrow bc = 0$. Similarly,$23$ pairs.Case $II$: $a = d$. Then $b(2a) = 0$ and $c(2a) = 0$. If $a 
eq 0$,then $b = c = 0$. Since $a^2 = 1$,$a = 1$ or $a = -1$. This gives $(1, 1)$ and $(-1, -1)$ for $(a, d)$. ($2$ pairs).If $a = 0$,then $d = 0$,which leads to $bc = 1$,already covered in Case $I$.Total = $2 + 23 + 23 + 2 = 50$.

The number of matrices $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,where $a, b, c, d \in \{-1, 0, 1, 2, 3, \ldots, 10\}$,such that $A=A^{-1}$,is

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