Determinant/Recursively/Alternating/Fact/Proof

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

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

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

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