Discussion
The matrix
\[ \begin{bmatrix} (3 - x) & 2 & 2 \\ 2 & (4 - x) & 1 \\ -2 & -4 & (-1 - x) \end{bmatrix} \]
is singular for the following values of \(x\)
Answer: A
The matrix is singular for \(x = 0\) and \(x = 3\).
The matrix is \(A - xI\) with \(A = \begin{bmatrix} 3 & 2 & 2 \\ 2 & 4 & 1 \\ -2 & -4 & -1 \end{bmatrix}\), so it is singular exactly when \(x\) is an eigenvalue of \(A\). Use the invariants of \(A\):
\[ \begin{aligned} \det A &= 3(0) - 2(0) + 2(0) = 0 \\ \text{trace} &= 3 + 4 - 1 = 6 \\ \text{sum of 2x2 minors} &= 8 + 1 + 0 = 9 \end{aligned} \]
The zero determinant gives one eigenvalue 0. The other two add to 6 and multiply to 9 (the sum of the principal minors equals \(\lambda_{1}\lambda_{2}\) when the third eigenvalue is 0), so both are 3. The characteristic equation is \(x(x - 3)^{2} = 0\), giving \(x = 0\) and \(x = 3\).
The values -3, 6 and -6 are not roots: 6 is the trace, the sum of all three eigenvalues, not one of them, and the negative values have the wrong sign.
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