<s>[INST] Let $A$ be a $3$X$3$ matrix whose eigenvalues are $1$, $2$, $3$. Find $\det(B)$ where $B = A^2 + A^T$. Let $A$ be a $3$X$3$ matrix whose eigenvalues are $1$, $2$, $3$. Find $\det(B)$ where $B = A^2 + A^T$.
I know that $\det(A) = 6$, but I cannot proceed after $|A^2 + A^T|$. Any hints as to how to approach the problem?
 [/INST] The answer, it seems, is that it depends.  If we have $$ A = \left[ \begin{array}{ccc}
1 & 0 & 1 \\ 
0 & 2 & 0 \\
0 & 0 & 3 
\end{array}
\right]$$
Then we have $\det(A^2 + A^T) = 120$.
On the other hand, if we have $$ A = \left[ \begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 2 & 0 \\
0 & 0 & 3 
\end{array}
\right]$$
Then we have $\det(A^2 + A^T) = 144$.
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