Another old idea. Given a $2 \times 2$ matrix $\mathbf{A}$,

$$\det(\mathbf{A}) = \begin{bmatrix}1&0\end{bmatrix}\mathbf{A}^T\begin{bmatrix}0&1\\-1&0\end{bmatrix}\mathbf{A}\begin{bmatrix}0\\1\end{bmatrix}.$$

Prove this and extend it to general $n \times n$ matrices. Can you go further than that?

$$\det(\mathbf{A}) = \begin{bmatrix}1&0\end{bmatrix}\mathbf{A}^T\begin{bmatrix}0&1\\-1&0\end{bmatrix}\mathbf{A}\begin{bmatrix}0\\1\end{bmatrix}.$$

Prove this and extend it to general $n \times n$ matrices. Can you go further than that?