关于高斯消元求解行列式的证明
\[
已知 det(A) =
\left|\begin{matrix}
a_{1,1} & a_{1,2} & ... & a_{1,n}\\
a_{2,1} & a_{2,2} & ... & a_{2,n}\\
... & ... & ... & ... \\
a_{n,1} & a_{n,2} & ... & a_{n,n}\\
\end{matrix} \right|
\]
\[
设det(B) =
\left|\begin{matrix}
k\times a_{2,1} & k\times a_{2,2} & ... & k\times a_{2,n}\\
1\times a_{2,1} & 1\times a_{2,2} & ... & 1\times a_{2,n}\\
... & ... & ... & ... \\
1\times a_{n,1} & 1\times a_{n,2} & ... & 1\times a_{n,n}\\
\end{matrix} \right|
\]
\[
设det(C)=det(A)+det(B)
\]
\[
det(C)=
\left|\begin{matrix}
a_{1,1}+k\times a_{2,1} & a_{1,2}+k\times a_{2,2} & ... & a_{1,n}+k\times a_{2,n}\\
1\times a_{2,1} & 1\times a_{2,2} & ... & 1\times a_{2,n}\\
... & ... & ... & ... \\
1\times a_{n,1} & 1\times a_{n,2} & ... & 1\times a_{n,n}\\
\end{matrix} \right|
\]
\[
因为det(B)的第一行和第二行成比例
\]
\[
所以det(B)=0,det(C)=det(A)+det(B)=det(A)
\]
\[
即可以用高斯消元求解行列式
\]