Question

If $A = \begin{bmatrix} k & 2 \\ -2 & -k \end{bmatrix}$, then $A^{-1}$ does not exist if $k =$

• A. $3$
• B. $\pm 2$
• C. $0$
• D. $\pm 1$

Hint:

For any $2 \times 2$ matrix of the form $\begin{bmatrix} \alpha & \beta
-\beta & -\alpha \end{bmatrix}$, the determinant is always equal to $-\alpha^2 + \beta^2$. Setting it to zero gives $\alpha^2 = \beta^2 \implies \alpha = \pm \beta$. Here, $k = \pm 2$ can be written down directly without expanding the full determinant algebra!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Find the values of k for which the inverse of matrix A = [[k, 2], [-2, -k]] does not exist.

Step 2: Key Formula or Approach:
A matrix is non-invertible (singular) exactly when its determinant equals zero. For a 2×2 matrix, det = ad - bc.

Step 3: Detailed Explanation:
Compute det(A) = (k)(-k) - (2)(-2) = -k² + 4. For A⁻¹ to not exist, set det = 0: -k² + 4 = 0 → k² = 4 → k = ±2.

Step 4: Final Answer:
The inverse fails to exist when k = ±2, option (B).