Formula for $(i,j)$-th cofactor

Question

Let $A$ be an $n \times n$ matrix and denote by $M_{ji}$ the $(j,i)$-th cofactor of $A$, namely $(-1)^{i+j} \det A[ij]$, where $A[ij]$ is obtained from $A$ by deleting the $i$-th row and $j$-th column. Does it hold that $$M_{ji} = \sum_{\sigma \in S_n, \sigma(i) = j} \text{sgn}(\sigma) \, A_{1 \sigma(1)} \cdot \dots \cdot \hat{A}_{ij} \cdot \dots \cdot A_{n \sigma(n)},$$ where the hat means omit that factor? Of course $$M_{ji} = (-1)^{i+j} \sum_{\tau \in S_{n-1}} \text{sgn}(\tau) \, A_{1 \tau(1)} \cdot \dots \cdot A_{n-1, \tau(n-1)};$$ since each sum contains $(n-1)!$ terms, I tried to construct a correspondence between $\tau$ and $\sigma$ for which the associated summands coincide: $$\sigma(k) = \begin{cases} \tau(k) &\quad \text{if} \quad {k < i, \tau(k) < j}\\ \tau(k-1) &\quad \text{if} \quad {k \ge i+1, \tau(k-1) < j}\\ j &\quad \text{if} \quad {k =i} \\ \tau(k)+1 &\quad \text{if} \quad {k < i, \tau(k) \ge j} \\ \tau(k-1)+1 &\quad \text{if} \quad {k \ge i+1, \tau(k-1) \ge j}.\\ \end{cases}$$ However, determining $\text{sgn}(\sigma)$ in terms of $\tau$ is a nightmare, so I'm not quite there. Thanks for any advice.

Answer 1

Hint:

You can define an index mapping function

$$M_n(m):=\begin{cases}m<n\to m\\m\ge n\to n+1\end{cases}$$ and write the formula for the determinant of the $n-1\times n-1$ matrix

$$B_{kl}:=A_{M_i(k)M_j(l)}.$$