Question:
If the matrix \(
\begin{pmatrix}
8-k & 2
-2 & 4-k
\end{pmatrix}
\) is singular, then the value of \( k \) is equal to
If the matrix \(
\begin{pmatrix}
8-k & 2
-2 & 4-k
\end{pmatrix}
\) is singular, then the value of \( k \) is equal to
Show Hint
For singular matrices, always set determinant equal to zero and solve.
Updated On: Apr 21, 2026
- \(6 \)
- \(5 \)
- \(4 \)
- \(3 \)
- \(2 \)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
A matrix is singular if determinant = 0.
Step 1: Compute determinant.
\[ \begin{vmatrix} 8-k & 2 -2 & 4-k \end{vmatrix} = (8-k)(4-k) - (-2)(2) \]
Step 2: Simplify.
\[ = (8-k)(4-k) + 4 \] \[ = (32 - 8k - 4k + k^2) + 4 \] \[ = k^2 - 12k + 36 \]
Step 3: Set determinant = 0.
\[ k^2 - 12k + 36 = 0 \] \[ (k-6)^2 = 0 \Rightarrow k = 6 \]
Step 1: Compute determinant.
\[ \begin{vmatrix} 8-k & 2 -2 & 4-k \end{vmatrix} = (8-k)(4-k) - (-2)(2) \]
Step 2: Simplify.
\[ = (8-k)(4-k) + 4 \] \[ = (32 - 8k - 4k + k^2) + 4 \] \[ = k^2 - 12k + 36 \]
Step 3: Set determinant = 0.
\[ k^2 - 12k + 36 = 0 \] \[ (k-6)^2 = 0 \Rightarrow k = 6 \]
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