Question

If \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), where \( a, b, c, d \in \{-1, 1\} \), then the number of singular matrices in \( A \) is:

• A. 9
• B. 12
• C. 10
• D. 8

Hint:

For singular matrices, check the condition \( ad - bc = 0 \) when \( a, b, c, d \) are chosen from specific sets of values.

Answer 1

The Correct Option is D

A matrix is singular if its determinant is zero. The determinant of a 2x2 matrix is given by: \[ \text{det}(A) = ad - bc \] We are given that \( a, b, c, d \in \{-1, 1\} \), so the possible values for \( a, b, c, d \) are all combinations of \( -1 \) and \( 1 \). For \( A \) to be singular, we need \( ad - bc = 0 \). This implies: \[ ad = bc \] Now, let’s check all the possible combinations of \( a, b, c, d \) where \( a, b, c, d \in \{-1, 1\} \): - \( a = 1, b = 1, c = 1, d = 1 \), then \( 1 \times 1 - 1 \times 1 = 0 \), so singular. - \( a = 1, b = 1, c = 1, d = -1 \), then \( 1 \times (-1) - 1 \times 1 = -2 \), not singular. - \( a = 1, b = 1, c = -1, d = 1 \), then \( 1 \times 1 - 1 \times (-1) = 2 \), not singular. - \( a = 1, b = 1, c = -1, d = -1 \), then \( 1 \times (-1) - 1 \times (-1) = 0 \), so singular. - Similarly, check other combinations. The valid combinations where \( ad - bc = 0 \) are: - \( (1, 1, 1, 1), (1, -1, 1, -1), (-1, 1, -1, 1), (-1, -1, -1, -1) \) - These yield 8 singular matrices in total. Thus, the correct answer is option (4), 8.