## The cofactor expansion formula of a square matrix along a column

Let $A$ be a square matrix over a field $F$. Formulate and prove the cofactor expansion formula for $\mathsf{det}\ A$ along the $j$th column.

Let $A = [\alpha_{i,j}]$.

We begin with a definition. If $A = \left[ \begin{array}{c|c|c} A_{1,1} & C_1 & A_{1,2} \\ \hline R_1 & \alpha_{i,j} & R_2 \\ \hline A_{2,1} & C_2 & A_{2,2} \end{array} \right]$, where $A_{1,1}$ has dimension $(i-1) \times (j-1)$, then the $(i,j)$-minor of $A$ is the matrix $A_{i,j} = \left[ \begin{array}{c|c} A_{1,1} & A_{1,2} \\ \hline A_{2,1} & A_{2,2} \end{array} \right]$.

Recall that the cofactor expansion formula for $\mathsf{det}\ A$ along the $i$th row is

$\mathsf{det}\ A = \sum_{j=1}^n (-1)^{i+j} \alpha_{i,j} \mathsf{det}(A_{i,j})$.

The analogous expansion along the $j$th column is

$\mathsf{det}\ A = \sum_{i=1}^n (-1)^{i+j} \alpha_{i,j} \mathsf{det}(A_{i,j})$.

which we presently prove to be true. First, note that $(A_{i,j})^\mathsf{T} = (A^\mathsf{T})_{j,i}$; this follows from our definition of minors and the fact that $\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^\mathsf{T} = \left[ \begin{array}{c|c} A^\mathsf{T} & C^\mathsf{T} \\ \hline B^\mathsf{T} & D^\mathsf{T} \end{array} \right]$. Now we have the following.

 $\mathsf{det}(A)$ = $\mathsf{det}(A^\mathsf{T})$ = $\sum_{j=1}^n (-1)^{i+j} \alpha_{j,i} \mathsf{det}((A^\mathsf{T})_{i,j})$ = $\sum_{j=1}^n (-1)^{i+j} \alpha_{j,i} \mathsf{det}((A_{j,i})^\mathsf{T})$ = $\sum_{j=1}^n (-1)^{i+j} \alpha_{j,i} \mathsf{det}(A_{j,i})$ = $\sum_{i=1}^n (-1)^{i+j} \alpha_{i,j} \mathsf{det}(A_{i,j})$,

as desired.