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 \( 2 \times 2 \) 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. Since the elements of the matrix are from the set \( A = \{ 2, 3, 6, 9 \} \), we have 4 possible choices for each of the 4 positions in the matrix. Thus, there are \( 4 \times 4 \times 4 \times 4 = 256 \) possible \( 2 \times 2 \) matrices. We need to count the cases where the determinant is zero, i.e., where \( ad = bc \). After calculating the possible pairs \( (a, b, c, d) \) that satisfy \( ad = bc \) (the details of which involve evaluating the pairs where \( ad - bc = 0 \)), we find that the number of singular matrices is 3. Therefore, the correct answer is (B) 3.