<s>[INST] finding the characteristic polynomial of symmetric matrix So I got this symmetric matrix
$$A=        \begin{pmatrix}
        0 & 3 & 1 & -2 \\
        3 & 0 & -2 & 1 \\
        1 & -2 & 0 & 3 \\
        -2 & 1 & 3 & 0\\
        \end{pmatrix}$$
I've got to $$P(t)=\begin{pmatrix}
        -t & 3 & 1 & -2 \\
        3 & -t & -2 & 1 \\
        1 & -2 & -t & 3 \\
        -2 & 1 & 3 & -t\\
        \end{pmatrix}.$$
My question is does the simple multiplication method not work for this? When I multiply the diagnols to get the determinant I end up with $t^4-2t^2-24$.
I've also tries $a_{i,j}*det\begin{pmatrix}
a&b&c\\
b&c&d\\
c&d&e\\
\end{pmatrix}$ method(does this have a name?), but still it didn't seem right.
When using the Gaussian elimination I get $t(t-2)(t-4)(t+6)$, which seems to be the correct answer.
I'm guessing the other ones doesn't meet the some condition here, But what is it? And is there any other way than using the gaussian elimination here?
 [/INST] The Laplace expansion with cofactors works:
$-t\begin{vmatrix}-t&-2&1\\-2&-t&3\\1&3&-t\end{vmatrix}-3\begin{vmatrix}3&-2&1\\1&-t&3\\-2&3&-t\end{vmatrix}+\begin{vmatrix}3&-t&1\\1&-2&3\\2&1&-t\end{vmatrix}+2\begin{vmatrix}3&-t&-2\\1&-2&-t\\-2&1&3\end{vmatrix}$
$=t^4 - 28 t^2 + 48 t=t(t - 4) (t - 2) (t + 6).$
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