(1)

\(\boldsymbol{a}_{m}=\left(a_{1,m},a_{2,m},\cdots a_{n,m}\right)^{T}\)とする。
\begin{align*} \det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{j},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right) & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots a_{\sigma\left(j\right),j}\cdots a_{\sigma\left(k\right),k}\cdots a_{\sigma\left(n\right),n}\\ & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\left(\sigma\left(j\right),\sigma\left(k\right)\right)\right)a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots a_{\sigma\left(k\right),j}\cdots a_{\sigma\left(j\right),k}\cdots a_{\sigma\left(n\right),n}\\ & =-\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots a_{\sigma\left(k\right),j}\cdots a_{\sigma\left(j\right),k}\cdots a_{\sigma\left(n\right),n}\\ & =-\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{j},\cdots,\boldsymbol{a}_{n}\right) \end{align*} となる。
また、
\begin{align*} \det\left(\boldsymbol{a}_{\sigma\left(1\right)},\boldsymbol{a}_{\sigma\left(2\right)},\cdots,\boldsymbol{a}_{\sigma\left(n\right)}\right) & =\sum_{\tau\in S_{n}}\sgn\left(\tau\right)a_{\tau\left(1\right),\sigma\left(1\right)},a_{\tau\left(2\right),\sigma\left(2\right)},\cdots,a_{\tau\left(n\right),\sigma\left(n\right)}\\ & =\sum_{\tau\in S_{n}}\sgn\left(\tau\right)a_{\left(\sigma^{\bullet}\tau\right)\left(1\right),1},a_{\left(\sigma^{\bullet}\tau\right)\left(2\right),2},\cdots,a_{\left(\sigma^{\bullet}\tau\right)\left(n\right),n}\\ & =\sgn\left(\sigma\right)\sum_{\tau\in S_{n}}\sgn\left(\sigma^{\bullet}\tau\right)a_{\left(\sigma^{\bullet}\tau\right)\left(1\right),1},a_{\left(\sigma^{\bullet}\tau\right)\left(2\right),2},\cdots,a_{\left(\sigma^{\bullet}\tau\right)\left(n\right),n}\\ & =\sgn\left(\sigma\right)\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{n}\right) \end{align*} となるので与式は成り立つ。

(2)

(1)より、\(A\)の同じ列(行))同士を入れ替えると符号が変わるが行列\(A\)はそのままなので、\(\det\left(A\right)=-\det\left(A\right)\)となり\(\det\left(A\right)=0\)となる。
従って題意は成り立つ。

(3)

\(\boldsymbol{a}_{m}=\left(a_{1,m},a_{2,m},\cdots a_{n,m}\right)^{T}\)とする。
\begin{align*} \det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{k}+\boldsymbol{b}_{k},\cdots,\boldsymbol{a}_{n}\right) & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots\left(a_{\sigma\left(k\right),k}+b_{\sigma\left(k\right),k}\right)\cdots a_{\sigma\left(n\right),n}\\ & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)\left(a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots a_{\sigma\left(k\right),k}\cdots a_{\sigma\left(n\right),n}+a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots b_{\sigma\left(k\right),k}\cdots a_{\sigma\left(n\right),n}\right)\\ & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots a_{\sigma\left(k\right),k}\cdots a_{\sigma\left(n\right),n}+\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots b_{\sigma\left(k\right),k}\cdots a_{\sigma\left(n\right),n}\\ & =\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right)+\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{b}_{k},\cdots,\boldsymbol{a}_{n}\right) \end{align*} となるので与式は成り立つ。

(4)

\(\boldsymbol{a}_{m}=\left(a_{1,m},a_{2,m},\cdots a_{n,m}\right)^{T}\)とする。
\begin{align*} \det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,c\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right) & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots ca_{\sigma\left(k\right),k}\cdots a_{\sigma\left(n\right),n}\\ & =c\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\sigma\left(1\right),1}a_{\sigma\left(2\right),2}\cdots a_{\sigma\left(k\right),k}\cdots a_{\sigma\left(n\right),n}\\ & =c\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right) \end{align*}

(5)

\begin{align*} \det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{j},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right) & =\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{j},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right)+c\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right)\\ & =\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{j},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right)+\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,c\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right)\\ & =\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{j}+c\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{k},\cdots,\boldsymbol{a}_{n}\right) \end{align*} となるので与式は成り立つ。

(6)

行列\(A\)は
\[ A=\left(\begin{array}{cccccc} a_{1,1} & a_{1,2} & \cdots & a_{1,j} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,j} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{i,1} & a_{i,2} & \cdots & a_{i,j} & \cdots & a_{i,n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,j} & \cdots & a_{n,n} \end{array}\right) \] として、第\(i\)行目は\(\left(i,j\right)\)成分以外が全て0なので、\(a_{i,1}=a_{i,2}=\cdots=a_{i,j-1}=a_{i,j+1}=\cdots=a_{i,n}=0\)である。
まず、行列\(A\)の\(\left(i,j\right)\)成分を行と列を入れ替えて、\(\left(n,n\right)\)の位置に移動させる。
このとき、行または列を1回入れ替えるごとに\(-1\)倍されるので、
\begin{align*} \det A & =\det\left(\begin{array}{cccccc} a_{1,1} & a_{1,2} & \cdots & a_{1,j} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,j} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{i,1} & a_{i,2} & \cdots & a_{i,j} & \cdots & a_{i,n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,j} & \cdots & a_{n,n} \end{array}\right)\\ & =\left(-1\right)^{n-i}\det\left(\begin{array}{cccccc} a_{1,1} & a_{1,2} & \cdots & a_{1,j} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,j} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{i-1,1} & a_{i-1,2} & \cdots & a_{i-1,j} & \cdots & a_{i-1,n}\\ a_{i+1,1} & a_{i+1,2} & \cdots & a_{i+1,j} & \cdots & a_{i+1,n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,j} & \cdots & a_{n,n}\\ a_{i,1} & a_{i,2} & \cdots & a_{i,j} & \cdots & a_{i,n} \end{array}\right)\\ & =\left(-1\right)^{n-i}\left(-1\right)^{n-j}\det\left(\begin{array}{cccccccc} a_{1,1} & a_{1,2} & \cdots & a_{1,j-1} & a_{1,j+1} & \cdots & a_{1,n} & a_{1,j}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,j-1} & a_{2,j+1} & \cdots & a_{2,n} & a_{2,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{i-1,1} & a_{i-1,2} & \cdots & a_{i-1,j-1} & a_{i-1,j+1} & \cdots & a_{i-1,n} & a_{i-1,j}\\ a_{i+1,1} & a_{i+1,2} & \cdots & a_{i+1,j-1} & a_{i+1,j+1} & \cdots & a_{i+1,n} & a_{i+1,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,j-1} & a_{n,j+1} & \cdots & a_{n,n} & a_{n,j}\\ a_{i,1} & a_{i,2} & \cdots & a_{i,j-1} & a_{i,j+1} & \cdots & a_{i,n} & a_{i,j} \end{array}\right)\\ & =\left(-1\right)^{i+j}\det\left(\begin{array}{cccccccc} a_{1,1} & a_{1,2} & \cdots & a_{1,j-1} & a_{1,j+1} & \cdots & a_{1,n} & a_{1,j}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,j-1} & a_{2,j+1} & \cdots & a_{2,n} & a_{2,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{i-1,1} & a_{i-1,2} & \cdots & a_{i-1,j-1} & a_{i-1,j+1} & \cdots & a_{i-1,n} & a_{i-1,j}\\ a_{i+1,1} & a_{i+1,2} & \cdots & a_{i+1,j-1} & a_{i+1,j+1} & \cdots & a_{i+1,n} & a_{i+1,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,j-1} & a_{n,j+1} & \cdots & a_{n,n} & a_{n,j}\\ a_{i,1} & a_{i,2} & \cdots & a_{i,j-1} & a_{i,j+1} & \cdots & a_{i,n} & a_{i,j} \end{array}\right) \end{align*} となる。
ここで、
\begin{align*} B & =\left(\begin{array}{cccc} b_{1,1} & b_{1,2} & \cdots & b_{1,n}\\ b_{2,1} & b_{2,2} & \cdots & b_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ b_{n,1} & b_{n,2} & \cdots & b_{n,n} \end{array}\right)\\ & =\left(\begin{array}{cccccccc} a_{1,1} & a_{1,2} & \cdots & a_{1,j-1} & a_{1,j+1} & \cdots & a_{1,n} & a_{1,j}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,j-1} & a_{2,j+1} & \cdots & a_{2,n} & a_{2,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{i-1,1} & a_{i-1,2} & \cdots & a_{i-1,j-1} & a_{i-1,j+1} & \cdots & a_{i-1,n} & a_{i-1,j}\\ a_{i+1,1} & a_{i+1,2} & \cdots & a_{i+1,j-1} & a_{i+1,j+1} & \cdots & a_{i+1,n} & a_{i+1,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,j-1} & a_{n,j+1} & \cdots & a_{n,n} & a_{n,j}\\ a_{i,1} & a_{i,2} & \cdots & a_{i,j-1} & a_{i,j+1} & \cdots & a_{i,n} & a_{i,j} \end{array}\right) \end{align*} とおく。
このとき、\(a_{i,1}=a_{i,2}=\cdots=a_{i,j-1}=a_{i,j+1}=\cdots=a_{i,n}=0\)なので\(b_{n,1}=b_{n,2}=\cdots=b_{n,n-1}=0\)となる。
そうすると、
\begin{align*} \det A & =\left(-1\right)^{i+j}\det\left(\begin{array}{cccccccc} a_{1,1} & a_{1,2} & \cdots & a_{1,j-1} & a_{1,j+1} & \cdots & a_{1,n} & a_{1,j}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,j-1} & a_{2,j+1} & \cdots & a_{2,n} & a_{2,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{i-1,1} & a_{i-1,2} & \cdots & a_{i-1,j-1} & a_{i-1,j+1} & \cdots & a_{i-1,n} & a_{i-1,j}\\ a_{i+1,1} & a_{i+1,2} & \cdots & a_{i+1,j-1} & a_{i+1,j+1} & \cdots & a_{i+1,n} & a_{i+1,j}\\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,j-1} & a_{n,j+1} & \cdots & a_{n,n} & a_{n,j}\\ a_{i,1} & a_{i,2} & \cdots & a_{i,j-1} & a_{i,j+1} & \cdots & a_{i,n} & a_{i,j} \end{array}\right)\\ & =\left(-1\right)^{i+j}\det\left(B\right)\\ & =\left(-1\right)^{i+j}\det\left(\begin{array}{cccc} b_{1,1} & b_{1,2} & \cdots & b_{1,n}\\ b_{2,1} & b_{2,2} & \cdots & b_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ b_{n,1} & b_{n,2} & \cdots & b_{n,n} \end{array}\right)\\ & =\left(-1\right)^{i+j}\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)b_{1,\sigma\left(1\right)}b_{2,\sigma\left(2\right)}\cdots b_{n,\sigma\left(n\right)}\\ & =\left(-1\right)^{i+j}b_{n,n}\sum_{\sigma\in S_{n},\sigma\left(n\right)=n}\sgn\left(\sigma\right)b_{1,\sigma\left(1\right)}b_{2,\sigma\left(2\right)}\cdots b_{n-1,\sigma\left(n-1\right)}\cmt{\because b_{n,1}=b_{n,2}=\cdots=b_{n,n-1}=0}\\ & =\left(-1\right)^{i+j}b_{n,n}\sum_{\sigma\in S_{n-1}}\sgn\left(\sigma\right)b_{1,\sigma\left(1\right)}b_{2,\sigma\left(2\right)}\cdots b_{n-1,\sigma\left(n-1\right)}\\ & =\left(-1\right)^{i+j}b_{n,n}\det\left(\begin{array}{cccc} b_{1,1} & b_{1,2} & \cdots & b_{1,n-1}\\ b_{2,1} & b_{2,2} & \cdots & b_{2,n-1}\\ \vdots & \vdots & \ddots & \vdots\\ b_{n-1,1} & b_{n-1,2} & \cdots & b_{n-1,n-1} \end{array}\right)\\ & =\left(-1\right)^{i+j}b_{n,n}\det B_{n,n}\\ & =\left(-1\right)^{i+j}a_{i,j}\det A_{i,j} \end{align*} となり与式は成り立つ。