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(D) matrix $M = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is singular if its determinant is zero,i.e.,$ad - bc = 0$,which implies $ad = bc$.Since

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$8$

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(D) matrix $M = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is singular if its determinant is zero,i.e.,$ad - bc = 0$,which implies $ad = bc$.Since $a, b, c, d \in \{-1, 1\}$,the possible values for $ad$ are $1 	imes 1 = 1$,$1 	imes (-1) = -1$,$(-1) 	imes 1 = -1$,and $(-1) 	imes (-1) = 1$.There are $2$ ways to get $ad = 1$ (i.e.,$(1, 1)$ or $(-1, -1)$) and $2$ ways to get $ad = -1$ (i.e.,$(1, -1)$ or $(-1, 1)$).Similarly,there are $2$ ways to get $bc = 1$ and $2$ ways to get $bc = -1$.For $ad = bc$,we have two cases:Case $1$: $ad = 1$ and $bc = 1$. The number of ways is $2 	imes 2 = 4$.Case $2$: $ad = -1$ and $bc = -1$. The number of ways is $2 	imes 2 = 4$.Total number of singular matrices $= 4 + 4 = 8$.

If $A = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{-1, 1\} \right\}$,then the number of singular matrices in $A$ is

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