Question

If the matrix \( \begin{pmatrix} 8-k & 2 -2 & 4-k \end{pmatrix} \) is singular, then the value of \( k \) is equal to

• A. \(6 \)
• B. \(5 \)
• C. \(4 \)
• D. \(3 \)
• E. \(2 \)

Hint:

For singular matrices, always set determinant equal to zero and solve.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A matrix is said to be "singular" if its determinant is equal to zero. A singular matrix does not have a multiplicative inverse.
Step 2: Key Formula or Approach:
For a 2x2 matrix \(\begin{pmatrix} a & b
c & d \end{pmatrix}\), the determinant is calculated as `ad - bc`. We need to set the determinant of the given matrix to zero and solve for `k`.
\[ \det\begin{pmatrix} 8-k & 2
-2 & 4-k \end{pmatrix} = 0 \] Step 3: Detailed Explanation:
Calculate the determinant of the given matrix: \[ (8-k)(4-k) - (2)(-2) \] Set the determinant equal to zero: \[ (8-k)(4-k) + 4 = 0 \] Expand the product: \[ 32 - 8k - 4k + k^2 + 4 = 0 \] Combine like terms to form a standard quadratic equation: \[ k^2 - 12k + 36 = 0 \] This quadratic equation is a perfect square trinomial. It can be factored as: \[ (k - 6)^2 = 0 \] Solving for `k`: \[ k - 6 = 0 \] \[ k = 6 \] Step 4: Final Answer:
The value of k is 6.