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.

Answer 2

Given:
Matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), where each of \( a, b, c, d \in \{-1, 1\} \)
We are to find the number of **singular matrices**, i.e., matrices with **determinant = 0**.

Step 1: Total number of possible matrices
Each of \( a, b, c, d \) can be either −1 or 1.
So total number of such 2×2 matrices is:
\[ 2 \times 2 \times 2 \times 2 = 16 \]

Step 2: Condition for singular matrix
A matrix is singular if its determinant is 0.
For a 2×2 matrix: \[ \det A = ad - bc = 0 \Rightarrow ad = bc \]
So we are to count how many combinations of \( a, b, c, d \in \{-1, 1\} \) satisfy \( ad = bc \).

Step 3: List and count valid combinations
Let’s consider all 16 combinations and count how many satisfy \( ad = bc \):
We loop through all 16 combinations manually:
1. \( a = 1, b = 1, c = 1, d = 1 \Rightarrow ad = 1, bc = 1 \) ✅
2. \( a = 1, b = 1, c = 1, d = -1 \Rightarrow ad = -1, bc = 1 \) ❌
3. \( a = 1, b = 1, c = -1, d = 1 \Rightarrow ad = 1, bc = -1 \) ❌
4. \( a = 1, b = 1, c = -1, d = -1 \Rightarrow ad = -1, bc = -1 \) ✅
5. \( a = 1, b = -1, c = 1, d = 1 \Rightarrow ad = 1, bc = -1 \) ❌
6. \( a = 1, b = -1, c = 1, d = -1 \Rightarrow ad = -1, bc = -1 \) ✅
7. \( a = 1, b = -1, c = -1, d = 1 \Rightarrow ad = 1, bc = 1 \) ✅
8. \( a = 1, b = -1, c = -1, d = -1 \Rightarrow ad = -1, bc = 1 \) ❌
9. \( a = -1, b = 1, c = 1, d = 1 \Rightarrow ad = -1, bc = 1 \) ❌
10. \( a = -1, b = 1, c = 1, d = -1 \Rightarrow ad = 1, bc = 1 \) ✅
11. \( a = -1, b = 1, c = -1, d = 1 \Rightarrow ad = -1, bc = -1 \) ✅
12. \( a = -1, b = 1, c = -1, d = -1 \Rightarrow ad = 1, bc = -1 \) ❌
13. \( a = -1, b = -1, c = 1, d = 1 \Rightarrow ad = -1, bc = -1 \) ✅
14. \( a = -1, b = -1, c = 1, d = -1 \Rightarrow ad = 1, bc = -1 \) ❌
15. \( a = -1, b = -1, c = -1, d = 1 \Rightarrow ad = -1, bc = 1 \) ❌
16. \( a = -1, b = -1, c = -1, d = -1 \Rightarrow ad = 1, bc = 1 \) ✅

✅ Valid cases: 1, 4, 6, 7, 10, 11, 13, 16 → total = 8

Final Answer:
\[ \boxed{8} \]