let $m$ be an $n \times n$ matrix (over $\mathbb{R}$,say) and for a permutation $\sigma \in S_n$ define the monomial: $$ P_\sigma(M) = \prod_{j=1}^n m_{j,\sigma(j)} $$ let $\tau$ be an odd permutation, and $A_n$ be the alternating subgroup of $S_n$. then the determinant of $M$ is given by: $$ |M|= \sum_{\sigma \in A_n} P_\sigma - \sum_{\sigma \in \tau A_n}P_\sigma $$ if $M$ is singular we have $$ \sum_{\sigma \in A_n} P_\sigma = \sum_{\sigma \in \tau A_n}P_\sigma $$
question is there any direct intuitive explanation of this fact?