linear algebra - Coefficients of characteristic polynomial of a $3times 3$ matrix

Let $A$ be a $3\times 3$ matrix over reals. Then its characteristic polynomial $\det(xI-A)$ is of the form $x^3+a_2x^2+a_1x+a_0$. It is well known that

$$-a_2=\mbox{trace}(A) \mbox{ and } -a_0=\det(A).$$
Note that these constants are expressed as functions of $A$ without referring to eigenvalues of $A$.

Q. What is interpretation of $a_1$ in terms of $A$ without considering its eigenvalues?

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This could be trivial, but I have never seen it.

Answer

If your matrix is:

$$A=\begin{bmatrix} A_{11} & A_{12} & A_{13}\\A_{21} & A_{22} & A_{23}\\A_{31} & A_{32} & A_{33}\end{bmatrix}$$
Then your $a_1$ is:
$$a_1=\det\left(\begin{bmatrix} A_{22} & A_{23}\\A_{32} & A_{33}\end{bmatrix}\right)+\det\left(\begin{bmatrix} A_{11} & A_{13}\\A_{31} & A_{33}\end{bmatrix}\right)+\det\left(\begin{bmatrix} A_{11} & A_{12}\\A_{21} & A_{22}\end{bmatrix}\right)$$
And I think it's called $a_1=\text{Tr}(\text{Adj}(A))$.
Calculating it is similar to calculating the determinant: You are calculating the sub-determinant( I'm not sure about the name) for $A_{11}$, $A_{22}$ and $A_{33}$, and you are summing them.