Determinant/Recursively/Alternating/Fact/Proof

Proof

We proof the statement by induction over ${}n$,  So suppose that ${}n\geq 2$ and set ${}M={\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}={\left(a_{ij}\right)}_{ij}$. Let ${}v_{r}$ and ${}v_{s}$ with ${}r<s$ be the relevant row. By definition, we have ${}\det M=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det M_{i}$. Due to the induction hypothesis, we have ${}\det M_{i}=0$ for ${}i\neq r,s$, because two rows coincide in these cases. Therefore,

${}\det M=(-1)^{r+1}a_{r1}\det M_{r}+(-1)^{s+1}a_{s1}\det M_{s}\,,$

where ${}a_{r1}=a_{s1}$. The matrices ${}M_{r}$ and ${}M_{s}$ consist in the same rows, however, the row ${}z=v_{r}=v_{s}$ is in ${}M_{r}$ the ${}(s-1)$-th row and in ${}M_{s}$ the ${}r$-th row. All other rows occur in both matrices in the same order. By swapping altogether ${}s-r-1$ times adjacent rows, we can transform ${}M_{r}$ into ${}M_{s}$. Due to the induction hypothesis and fact, their determinants are related by the factor ${}(-1)^{s-r-1}$, thus ${}\det M_{s}=(-1)^{s-r-1}\det M_{r}$. Using this, we obtain

${}{\begin{aligned}\det M&=(-1)^{r+1}a_{r1}\det M_{r}+(-1)^{s+1}a_{s1}\det M_{s}\\&=a_{r1}{\left((-1)^{r+1}\det M_{r}+(-1)^{s+1}(-1)^{s-r-1}\det M_{r}\right)}\\&=a_{r1}{\left((-1)^{r+1}+(-1)^{2s-r}\right)}\det M_{r}\\&=a_{r1}{\left((-1)^{r+1}+(-1)^{r}\right)}\det M_{r}\\&=0.\end{aligned}}$

To fact