Let \( A \) be a set defined as \( A = \{ 2, 3, 6, 9 \} \). Find the number of singular matrices of order \( 2 \times 2 \) such that elements are from the set \( A \).
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A matrix is singular if its determinant is zero. For a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b c & d \end{pmatrix} \), the determinant is \( ad - bc \).
A \( 2 \times 2 \) matrix is singular if its determinant is zero. For a \( 2 \times 2 \) matrix, the determinant is given by:\[\text{det}(A) = \begin{vmatrix} a & b c & d \end{vmatrix} = ad - bc\]The objective is to determine the count of matrices with a determinant of zero. Given that the matrix elements are drawn from the set \( A = \{ 2, 3, 6, 9 \} \), there are 4 options for each of the 4 entries in the matrix.Consequently, the total number of possible \( 2 \times 2 \) matrices is \( 4 \times 4 \times 4 \times 4 = 256 \). The requirement is to enumerate the instances where the determinant equals zero, meaning \( ad = bc \).Upon computation of the pairs \( (a, b, c, d) \) satisfying \( ad = bc \), (details involving enumeration of pairs where \( ad - bc = 0 \)), it is found that 3 matrices are singular.Hence, the answer is (B) 3.
