Discussion
One eigenvalue of the matrix \(A = \begin{bmatrix} 2 & 7 & 10 \\ 5 & 2 & 25 \\ 1 & 6 & 5 \end{bmatrix}\) is -9.33. One of the other eigenvalues is
Answer: A
Another eigenvalue is 18.33.
The determinant of A is zero:
\[ \begin{aligned} |A| &= 2(10 - 150) - 7(25 - 25) + 10(30 - 2) \\ &= -280 - 0 + 280 = 0 \end{aligned} \]
so one eigenvalue is 0. The eigenvalues add up to the trace, \(2 + 2 + 5 = 9\). With 0 and -9.33, the third is \(9 - 0 + 9.33 = 18.33\). As a check, the sum of the principal 2 x 2 minors, \(-31 + 0 - 140 = -171\), equals \(18.33 \times (-9.33) \approx -171\).
A real matrix has complex eigenvalues only in conjugate pairs, so a lone \(18.33 \pm 9.33i\) cannot sit beside the real eigenvalues; and -18.33 would make the trace wrong.
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