Let M=left{A=begin{bmatrix} a & b c & d end{ | vedclass

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(A) We are given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ where $a, b, c, d \in \{\pm 3, \pm 2, \pm 1, 0\}$.We need to find the number of ma

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$16$

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(A) We are given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ where $a, b, c, d \in \{\pm 3, \pm 2, \pm 1, 0\}$.We need to find the number of matrices such that $\det(A) = ad - bc = 15$.Since the maximum value of $ad$ is $3 	imes 3 = 9$ and the minimum value of $bc$ is $-3 	imes 3 = -9$,the maximum value of $ad - bc$ is $9 - (-9) = 18$.Possible values for $ad$ and $bc$ such that $ad - bc = 15$:Case $I$: $ad = 9$ and $bc = -6$.For $ad = 9$,the pairs $(a, d)$ can be $(3, 3)$ or $(-3, -3)$ ($2$ pairs).For $bc = -6$,the pairs $(b, c)$ can be $(3, -2), (-3, 2), (-2, 3), (2, -3)$ ($4$ pairs).Total matrices for Case $I = 2 	imes 4 = 8$.Case $II$: $ad = 6$ and $bc = -9$.For $ad = 6$,the pairs $(a, d)$ can be $(3, 2), (2, 3), (-3, -2), (-2, -3)$ ($4$ pairs).For $bc = -9$,the pairs $(b, c)$ can be $(3, -3), (-3, 3)$ ($2$ pairs).Total matrices for Case $II = 4 	imes 2 = 8$.Total number of such matrices $= 8 + 8 = 16$.

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