Discussion
If the matrix \(\begin{bmatrix} 8 - k & 2 \\ -2 & 4 - k \end{bmatrix}\) is singular, then the value of \(k\) is equal to
Answer: A
\(k = \) 6.
A matrix is singular when its determinant is zero:
\[ \begin{aligned} (8 - k)(4 - k) - (2)(-2) &= 0 \\ k^{2} - 12k + 32 + 4 &= 0 \\ (k - 6)^{2} &= 0 \end{aligned} \]
So \(k = 6\) (a repeated root). Check: the matrix becomes \(\begin{bmatrix} 2 & 2 \\ -2 & -2 \end{bmatrix}\), whose rows are proportional. Subtracting 4 instead of adding it gives \(k^{2} - 12k + 28 = 0\), which has no integer root among 2 to 5.
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