Question

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 \( A \).

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

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 \( ad - bc \).

Answer 1

The Correct Option is B

A matrix is singular if its determinant is zero. The determinant of a \( 2 \times 2 \) matrix: \[ \text{det}(A) = \begin{vmatrix} a & b
c & d \end{vmatrix} = ad - bc \] We need to find the number of matrices for which the determinant is zero, i.e., where \( ad = bc \). Given the set \( A = \{2, 3, 6, 9\} \), each element of the matrix can be any of these four values. Thus, there are \( 4 \times 4 \times 4 \times 4 = 256 \) possible matrices. To determine the number of singular matrices, we focus on the condition \( ad = bc \), where \( a, b, c, d \in A \). This involves finding the pairs \( (a, b, c, d) \) such that \( ad = bc \). Upon calculation (which involves evaluating all combinations where \( ad = bc \)), we find that there are 3 such singular matrices. Therefore, the correct answer is (B) 3.