Question

If $$ A = \begin{pmatrix} k + 1 & 2 \\4 & k - 1 \end{pmatrix}$$ is a singular matrix, then possible values of k are

• A. ±1
• B. ±2
• C. ±3
• D. ±4

Hint:

A matrix is singular if its determinant is zero. For a 2x2 matrix \( \begin{pmatrix} a & b \\c & d \end{pmatrix} \), the determinant is calculated as \( \text{det}(A) = ad - bc \).

Answer 1

The Correct Option is C

To determine the values of \( k \) that make matrix \( A \) singular:

Given Matrix:
\[ A = \begin{bmatrix} k+1 & 2 \\ 4 & k-1 \end{bmatrix} \]

1. Singularity Condition:
A matrix is singular when its determinant equals zero: \[ |A| = 0 \]

2. Compute the Determinant:
\[ |A| = (k+1)(k-1) - (4)(2) \]

3. Expand and Simplify:
\[ |A| = k^2 - 1 - 8 = k^2 - 9 \]

4. Solve the Equation:
Set the determinant to zero and solve for \( k \): \[ k^2 - 9 = 0 \implies k^2 = 9 \implies k = \pm 3 \]

Final Solution:
The matrix \( A \) is singular when: \[ k = 3 \quad \text{or} \quad k = -3 \]

Answer 2

A matrix is said to be singular if its determinant is zero. For the matrix \( A \), the determinant is given by: \[ \text{det}(A) = (k+1)(k-1) - (2)(4) \] Expanding this: \[ \text{det}(A) = k^2 - 1 - 8 = k^2 - 9 \] For the matrix to be singular, we set the determinant to zero: \[ k^2 - 9 = 0 \] Solving for \( k \): \[ k^2 = 9 \] \[ k = \pm 3 \]