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:
We are given a square $2 \times 2$ matrix $A$ containing a variable $k$. We need to find the value(s) of $k$ for which the multiplicative inverse matrix $A^{-1}$ fails to exist.

Step 2: Key Formula or Approach:
A square matrix is non-invertible (its inverse does not exist) if and only if it is a singular matrix. This means its determinant must be exactly equal to zero: $$\det(A) = 0$$ For a general $2 \times 2$ matrix $\begin{bmatrix} a & b
c & d \end{bmatrix}$, the determinant is calculated as $ad - bc$.

Step 3: Detailed Explanation:
Let's set up the determinant equation for the given matrix $A$: $$\det(A) = \begin{vmatrix} k & 2
-2 & -k \end{vmatrix}$$ Evaluating the determinant: $$\det(A) = (k)(-k) - (2)(-2)$$ $$\det(A) = -k^2 - (-4) = -k^2 + 4$$ For $A^{-1}$ to not exist, we set this expression to zero: $$-k^2 + 4 = 0$$ $$k^2 = 4$$ Taking the square root on both sides: $$k = \pm 2$$

Step 4: Final Answer:
The inverse matrix does not exist when $k = \pm 2$, which corresponds to option (B).