Discussion
If \(A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\), the characteristic roots of A are :
Answer: B
The characteristic roots are (1, 3).
Solve \(\det(A - \lambda I) = 0\):
\[ \begin{aligned} (2 - \lambda)^{2} - 1 &= 0 \\ 2 - \lambda &= \pm 1 \\ \lambda &= 1 \text{ or } 3 \end{aligned} \]
Check: the roots add to the trace, \(1 + 3 = 4 = 2 + 2\), and multiply to the determinant, \(1 \times 3 = 3 = 4 - 1\). The negative roots fail both checks, and (1, 1) has the wrong sum.
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