Here is a (requested) generalization for larger dimensions: If A is an n×n matrix, then one has the following expression for the determinant of A in terms of the determinant of a matrix whose entries are traces of powers of A:
$$\newcommand{\tr}{\operatorname{tr}}
\newcommand{\det}{\operatorname{det}}
\det(A) = \frac{1}{n!}\left|\begin{array}{cccccccc}
\tr(A) & 1 & . & . & . & \dots & . & . \\
\tr(A^2) & \tr(A) & 2 & . & . & \dots & . & . \\
\tr(A^3) & \tr(A^2) & \tr(A) & 3 & . & \dots & . & . \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
\tr(A^{n-1}) & \tr(A^{n-2}) & \tr(A^{n-3}) & \tr(A^{n-4}) & \tr(A^{n-5}) & \dots & \tr(A) & n-1 \\
\tr(A^n) & \tr(A^{n-1}) & \tr(A^{n-2}) & \tr(A^{n-3}) & \tr(A^{n-4}) & \dots & \tr(A^2) & \tr(A)
\end{array}\right|$$
In particular, we have for $3\times3$ that:
$$\det(A) = \frac{1}{6}\left|\begin{array}{cccc}
\tr(A) & 1 & 0 \\
\tr(A^2) & \tr(A) & 2 \\
\tr(A^3) & \tr(A^2) & \tr(A) \end{array}\right|$$
The matrix on the right is defined in general by: $$B_{ij} = \begin{cases}
i & \text{ if } j = i + 1 \\
\tr(A^{i-j+1}) & \text{ if } j \leq i \\
0 & \text{ if } j > i+1
\end{cases}$$
I suggest using this recursively to create even more complicated formulas (all in lower Hessenberg form!) to be entered into the big book of bad algorithms.