78. 杂例

例 78.1. 设 $A$ 是 $n$ 级阵. 设 $n$ 级阵 $D$ 适合$$\begin{array}{r}[D{]}_{i,j}=\{\begin{array}{ll}{d}_i,& i=j;\\ 0,& \text{其他}.\end{array}\end{array}$$则$$\begin{array}{rl}& \det (A+D)\\ =& \det (D)\\ & +\sum _{k=1}^{n-1}\sum _{1⩽j_1<j_2<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det (D(j_1,\dots ,j_k|j_1,\dots ,j_k))\\ & +\det (A).\end{array}$$

设 $A$ 的列 $1$, $2$, $\dots$, $n$ 分别是 $b_1^{(1)}$, $b_2^{(1)}$, $\dots$, $b_n^{(1)}$. 设 $D$ 的列 $1$, $2$, $\dots$, $n$ 分别是 $b_1^{(0)}$, $b_2^{(0)}$, $\dots$, $b_n^{(0)}$. 则 $A+D$ 的列 $j$ 是$$\begin{array}{r}{b}_j^{(1)}+b_j^{(0)}=b_j^{(0)}+b_j^{(1)}=\sum _{c_j=0}^1{b}_j^{(c_j)}.\end{array}$$则, 由多线性, $$\begin{array}{rl}& \det (A+D)\\ =& \det \left[\sum _{c_1=0}^1{b}_1^{(c_1)},\sum _{c_2=0}^1{b}_2^{(c_2)},\dots ,\sum _{c_n=0}^1{b}_n^{(c_n)}\right]\\ =& \sum _{c_1=0}^1\det \left[b_1^{(c_1)},\sum _{c_2=0}^1{b}_2^{(c_2)},\dots ,\sum _{c_n=0}^1{b}_n^{(c_n)}\right]\\ =& \sum _{c_1=0}^1\sum _{c_2=0}^1\det \left[b_1^{(c_1)},b_2^{(c_2)},\dots ,\sum _{c_n=0}^1{b}_n^{(c_n)}\right]\\ =& \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \\ =& \sum _{c_1=0}^1\sum _{c_2=0}^1\cdots \sum _{c_n=0}^1\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \sum _{0⩽c_1,c_2,\dots ,c_n⩽1}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}].\end{array}$$由加法的结合律与交换律, 我们可按任何方式, 任何次序求这些$$\begin{array}{r}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\end{array}$$的和. 特别地, 我们可按 $c_1+c_2+\cdots +c_n$ 分这些数为若干组, 求出一组的元的和 (即一组的 “组和”), 再求这些组的 “组和” 的和. 因为 $0⩽c_j⩽1$, 故 $0⩽c_1+c_2+\cdots +c_n⩽n$. 于是, 我们可分这些数为 $n+1$ 组: 适合 $c_1+c_2+\cdots +c_n=0$ 的项在一组; 适合 $c_1+c_2+\cdots +c_n=1$ 的项在一组; ……; 适合 $c_1+c_2+\cdots +c_n=n$ 的项在一组. 不难看出, 每一项在某一组里, 且每一项不能在二个不同的组里. 则$$\begin{array}{rl}& \det (A+D)\\ =& \sum _{0⩽c_1,c_2,\dots ,c_n⩽1}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \sum _{k=0}^n\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=k\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=0\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ & +\sum _{k=1}^{n-1}\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=k\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ & +\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=n\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}].\end{array}$$

不难看出, $c_1+c_2+\cdots +c_n=0$ 时, $c_1=c_2=\cdots =c_n=0$. 则$$\begin{array}{r}\sum _{0⩽c_1,c_2,\dots ,c_n⩽1}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]=\det [b_1^{(0)},b_2^{(0)},\dots ,b_n^{(0)}]=\det (D).\end{array}$$

不难看出, $c_1+c_2+\cdots +c_n=n$ 时, $c_1=c_2=\cdots =c_n=1$. 则$$\begin{array}{r}\sum _{0⩽c_1,c_2,\dots ,c_n⩽1}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]=\det [b_1^{(1)},b_2^{(1)},\dots ,b_n^{(1)}]=\det (A).\end{array}$$

设正整数 $k<n$. 由 $c_1+c_2+\cdots +c_n=k$, 知在 $c_1$, $c_2$, $\dots$, $c_n$ 中, 有 $k$ 个 $1$ 与 $n-k$ 个 $0$. 设 $c_{j_1}=c_{j_2}=\cdots =c_{j_k}=1$ (其中 $1⩽j_1<j_2<\cdots <j_k⩽n$), 且 $c_{j_{k+1}}=\cdots =c_{j_n}=0$ (其中 $1⩽j_{k+1}<\cdots <j_n⩽n$). 记 $n$ 级阵$$\begin{array}{r}{B}_{c_1,\dots ,c_n}=[b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}].\end{array}$$注意到, 当 $\ell >k$, 且 $i\ne j_{\ell }$ 时, $[B_{c_1,\dots ,c_n}{]}_{i,j_{\ell }}=0$, 且 $[B_{c_1,\dots ,c_n}{]}_{j_{\ell },j_{\ell }}=d_{j_{\ell }}$, 按列 $j_{k+1}$ 展开 $\det (B_{c_1,\dots ,c_n})$, 有$$\begin{array}{rl}\det (B_{c_1,\dots ,c_n})=& (-1{)}^{j_{k+1}+j_{k+1}}[B_{c_1,\dots ,c_n}{]}_{j_{\ell },j_{\ell }}\det (B_{c_1,\dots ,c_n}(j_{k+1}|j_{k+1}))\\ =& d_{j_{k+1}}\det (B_{c_1,\dots ,c_n}(j_{k+1}|j_{k+1})).\end{array}$$按列 $j_{k+2}-1$ 展开 $\det (B_{c_1,\dots ,c_n}(j_{k+1}|j_{k+1}))$, 有$$\begin{array}{rl}& \det (B_{c_1,\dots ,c_n}(j_{k+1}|j_{k+1}))\\ =& (-1{)}^{j_{k+2}-1+j_{k+2}-1}[B_{c_1,\dots ,c_n}{]}_{j_{\ell },j_{\ell }}\det (B_{c_1,\dots ,c_n}(j_{k+1},j_{k+2}|j_{k+1},j_{k+2}))\\ =& d_{j_{k+2}}\det (B_{c_1,\dots ,c_n}(j_{k+1},j_{k+2}|j_{k+1},j_{k+2})).\end{array}$$故$$\begin{array}{r}\det (B_{c_1,\dots ,c_n})=d_{j_{k+1}}{d}_{j_{k+2}}\det (B_{c_1,\dots ,c_n}(j_{k+1},j_{k+2}|j_{k+1},j_{k+2})).\end{array}$$…… 最后, 我们算出$$\begin{array}{rl}& \det (B_{c_1,\dots ,c_n})\\ =& d_{j_{k+1}}{d}_{j_{k+2}}\dots d_{j_n}\det (B_{c_1,\dots ,c_n}(j_{k+1},\dots ,j_n|j_{k+1},\dots ,j_n))\\ =& \det (B_{c_1,\dots ,c_n}(j_{k+1},\dots ,j_n|j_{k+1},\dots ,j_n))d_{j_{k+1}}{d}_{j_{k+2}}\dots d_{j_n}\\ =& \det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det (D(j_1,\dots ,j_k|j_1,\dots ,j_k)).\end{array}$$于是$$\begin{array}{rl}& \sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=k\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \sum _{1⩽j_1<j_2<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det (D(j_1,\dots ,j_k|j_1,\dots ,j_k)).\end{array}$$

综上, $$\begin{array}{rl}& \det (A+D)\\ =& \sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=0\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ & +\sum _{k=1}^{n-1}\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=k\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ & +\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=n\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \det (D)\\ & +\sum _{k=1}^{n-1}\sum _{1⩽j_1<j_2<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det (D(j_1,\dots ,j_k|j_1,\dots ,j_k))\\ & +\det (A).\end{array}$$

例 78.2. 设 $A$ 是 $n$ 级阵. 设 $x$ 是数. 则$$\begin{array}{r}\det (x{I}_n+A)=x^n+\sum _{k=1}^n{x}^{n-k}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right).\end{array}$$

注意到$$\begin{array}{r}[x{I}_n{]}_{i,j}=\{\begin{array}{ll}x,& i=j;\\ 0,& \text{其他}.\end{array}\end{array}$$故, 由上个例, $$\begin{array}{rl}& \det (x{I}_n+A)\\ =& \det (x{I}_n)\\ & +\sum _{k=1}^{n-1}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det ((x{I}_n)(j_1,\dots ,j_k|j_1,\dots ,j_k))\\ & +\det (A)\\ =& x^n+\sum _{k=1}^{n-1}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)x^{n-k}\\ & +\det (A)\\ =& x^n+\sum _{k=1}^{n-1}{x}^{n-k}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ & +x^{n-n}\sum _{1⩽j_1<\cdots <j_n⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_n}{j_1,\dots ,j_n}\right)\right)\\ =& x^n+\sum _{k=1}^n{x}^{n-k}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right).\end{array}$$

例 78.3. 设正整数 $m$, $n$ 适合 $m⩾n$. 设 $A$, $B$ 分别是 $m\times n$ 与 $n\times m$ 阵. 设正整数 $k⩽n$. 由 Binet–Cauchy 公式的推广, $$\begin{array}{rl}& \sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& \sum _{1⩽j_1<\cdots <j_k⩽m}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{i_1,\dots ,i_k}\right)\right)\det \left(B\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{j_1,\dots ,j_k}\right)\right)\\ =& \sum _{1⩽j_1<\cdots <j_k⩽m}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left(B\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{j_1,\dots ,j_k}\right)\right)\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{i_1,\dots ,i_k}\right)\right)\\ =& \sum _{1⩽i_1<\cdots <i_k⩽n}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left(B\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{j_1,\dots ,j_k}\right)\right)\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{i_1,\dots ,i_k}\right)\right)\\ =& \sum _{1⩽i_1<\cdots <i_k⩽n}\det \left((BA)\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{i_1,\dots ,i_k}\right)\right).\end{array}$$

例 78.4. 设正整数 $m$, $n$ 适合 $m⩾n$. 设 $A$, $B$ 分别是 $m\times n$ 与 $n\times m$ 阵. 设 $x$ 是数. 则$$\begin{array}{rl}& \det (x{I}_m+AB)\\ =& x^m+\sum _{k=1}^m{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& x^m+\sum _{k=1}^n{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ & +\sum _{k=n+1}^m{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& x^m+\sum _{k=1}^n{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ & +\sum _{k=n+1}^m{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}0\\ =& x^m+\sum _{k=1}^n{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& x^{m-n}{x}^n+\sum _{k=1}^n{x}^{m-n}{x}^{n-k}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left((BA)\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{i_1,\dots ,i_k}\right)\right)\\ =& x^{m-n}{x}^n+x^{m-n}\sum _{k=1}^n{x}^{n-k}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left((BA)\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{i_1,\dots ,i_k}\right)\right)\\ =& x^{m-n}\left(x^n+\sum _{k=1}^n{x}^{n-k}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left((BA)\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{i_1,\dots ,i_k}\right)\right)\right)\\ =& x^{m-n}\det (x{I}_n+BA).\end{array}$$特别地, 代 $x$ 以数 $1$, 有$$\begin{array}{r}\det (I_m+AB)=\det (I_n+BA).\end{array}$$

例 78.5. 设 $a_1$, $b_1$, $a_2$, $b_2$, $\dots$, $a_m$, $b_m$ 是 $2m$ 个数. 作 $m$ 级阵 $C$ 如下: $$\begin{array}{r}[C{]}_{i,j}=\{\begin{array}{ll}1+a_i{b}_i,& i=j;\\ a_i{b}_j,& \text{其他}.\end{array}\end{array}$$我们计算 $\det (C)$.

设 $A=[a_1,a_2,\dots ,a_m{]}^{\mathrm{T}}$, $B=[b_1,b_2,\dots ,b_m]$. 则$$\begin{array}{r}[AB{]}_{i,j}=[A{]}_{i,1}[B{]}_{1,j}=a_i{b}_j.\end{array}$$由此, 不难看出, $C=I_m+AB$. 故$$\begin{array}{rl}\det (C)=& \det (I_m+AB)\\ =& \det (I_1+BA)\\ =& [I_1+BA{]}_{1,1}\\ =& [I_1{]}_{1,1}+[BA{]}_{1,1}\\ =& 1+b_1{a}_1+b_2{a}_2+\cdots +b_m{a}_m.\end{array}$$

我们当然也可用别的方法. 作 $m+1$ 级阵 $G$ 如下: $$\begin{array}{r}[G{]}_{i,j}=\{\begin{array}{ll}1,& i=j=m+1;\\ -a_i,& i<j=m+1;\\ 0,& m+1=i>j;\\ [C{]}_{i,j},& \text{其他}.\end{array}\end{array}$$形象地, $$\begin{array}{r}G=\left[\begin{array}{cccccc}1+a_1{b}_1& a_1{b}_2& \cdots & a_1{b}_{m-1}& a_1{b}_m& -a_1\\ a_2{b}_1& 1+a_2{b}_2& \cdots & a_2{b}_{m-1}& a_2{b}_m& -a_2\\ ⋮& ⋮& & ⋮& ⋮& ⋮\\ a_{m-1}{b}_1& a_{m-1}{b}_2& \cdots & 1+a_{m-1}{b}_{m-1}& a_{m-1}{b}_m& -a_{m-1}\\ a_m{b}_1& a_m{b}_2& \cdots & a_m{b}_{m-1}& 1+a_m{b}_m& -a_m\\ 0& 0& \cdots & 0& 0& 1\end{array}\right].\end{array}$$一方面, 按行 $m+1$ 展开, 有$$\begin{array}{r}\det (G)=(-1{)}^{m+1+m+1}1\det (G(1|1))=\det (C).\end{array}$$另一方面, 我们加列 $m+1$ 的 $b_1$ 倍于列 $1$, 加列 $m+1$ 的 $b_2$ 倍于列 $2$, ……, 加列 $m+1$ 的 $b_m$ 倍于列 $m$, 得 $m+1$ 级阵$$\begin{array}{r}H=\left[\begin{array}{cccccc}1& 0& \cdots & 0& 0& -a_1\\ 0& 1& \cdots & 0& 0& -a_2\\ ⋮& ⋮& & ⋮& ⋮& ⋮\\ 0& 0& \cdots & 1& 0& -a_{m-1}\\ 0& 0& \cdots & 0& 1& -a_m\\ b_1& b_2& \cdots & b_{m-1}& b_m& 1\end{array}\right].\end{array}$$则 $\det (H)=\det (G)$. 故 $\det (C)=\det (H)$.

我们加列 $1$ 的 $a_1$ 倍于列 $m+1$, 加列 $1$ 的 $a_2$ 倍于列 $m+1$, ……, 加列 $1$ 的 $a_m$ 倍于列 $m+1$, 得 $m+1$ 级阵$$\begin{array}{r}J=\left[\begin{array}{cccccc}1& 0& \cdots & 0& 0& 0\\ 0& 1& \cdots & 0& 0& 0\\ ⋮& ⋮& & ⋮& ⋮& ⋮\\ 0& 0& \cdots & 1& 0& 0\\ 0& 0& \cdots & 0& 1& 0\\ b_1& b_2& \cdots & b_{m-1}& b_m& 1+b_1{a}_1+b_2{a}_2+\cdots +b_m{a}_m\end{array}\right].\end{array}$$则 $\det (J)=\det (H)$. 故$$\begin{array}{r}\det (C)=\det (J)=1+b_1{a}_1+b_2{a}_2+\cdots +b_m{a}_m.\end{array}$$

例 78.6. 设 $n$, $m$ 是非负整数, 且 $m+n⩾1$. 设$$\begin{array}{rl}f(x)& =\sum _{k=0}^n{a}_k{x}^{n-k}=a_0{x}^n+a_1{x}^{n-1}+\cdots +a_n,\\ g(x)& =\sum _{k=0}^m{b}_k{x}^{m-k}=b_0{x}^m+b_1{x}^{m-1}+\cdots +b_m.\end{array}$$为方便, 我们约定: 若 $k<0$ 或 $k>n$, 则 $a_k=0$; 若 $k<0$ 或 $k>m$, 则 $b_k=0$. 作 $m+n$ 级阵 $R$ 如下: $$\begin{array}{r}[R{]}_{i,j}=\{\begin{array}{ll}{a}_{i-j},& j⩽m;\\ b_{i-(j-m)},& j>m.\end{array}\end{array}$$形象地, 比如, 若 $n=2$, $m=5$, 则$$\begin{array}{r}R=\left[\begin{array}{ccccccc}{a}_0& a_{-1}& a_{-2}& a_{-3}& a_{-4}& b_0& b_{-1}\\ a_1& a_0& a_{-1}& a_{-2}& a_{-3}& b_1& b_0\\ a_2& a_1& a_0& a_{-1}& a_{-2}& b_2& b_1\\ a_3& a_2& a_1& a_0& a_{-1}& b_3& b_2\\ a_4& a_3& a_2& a_1& a_0& b_4& b_3\\ a_5& a_4& a_3& a_2& a_1& b_5& b_4\\ a_6& a_5& a_4& a_3& a_2& b_6& b_5\end{array}\right]=\left[\begin{array}{ccccccc}{a}_0& 0& 0& 0& 0& b_0& 0\\ a_1& a_0& 0& 0& 0& b_1& b_0\\ a_2& a_1& a_0& 0& 0& b_2& b_1\\ 0& a_2& a_1& a_0& 0& b_3& b_2\\ 0& 0& a_2& a_1& a_0& b_4& b_3\\ 0& 0& 0& a_2& a_1& b_5& b_4\\ 0& 0& 0& 0& a_2& 0& b_5\end{array}\right].\end{array}$$

(1) 设 $e$ 是 $m+n$ 级单位阵的列 $m+n$. 我们说, 存在 $(m+n)\times 1$ 阵 $p$ 使 $Rp=\det (R)e$.

取 $p$ 为 $R$ 的古伴 $\mathrm{adj}(R)$ 的列 $m+n$. 则$$\begin{array}{rl}[Rp{]}_{i,1}=& \sum _{\ell =1}^{m+n}[R{]}_{i,\ell }[p{]}_{\ell ,1}\\ =& \sum _{\ell =1}^{m+n}[R{]}_{i,\ell }[\mathrm{adj}(R){]}_{\ell ,m+n}\\ =& [R\mathrm{adj}(R){]}_{i,m+n}\\ =& [\det (R)I_{m+n}{]}_{i,m+n}\\ =& \det (R)[I_{m+n}{]}_{i,m+n}\\ =& \det (R)[e{]}_{i,1}\\ =& [\det (R)e{]}_{i,1}.\end{array}$$(顺便一提, 若 $\det (R)\ne 0$, 显然有 $p\ne 0$; 若 $\det (R)=0$, 由第一章, 节 23 的知识, 存在非零的 $(m+n)\times 1$ 阵 $q$ 使 $Rq=0=\det (R)e$.)

(2) 作 $1\times (m+n)$ 阵 $X=[x^{m+n-1},x^{m+n-2},\dots ,1]$; 具体地, $[X{]}_{1,j}=x^{m+n-j}$. 则 $XR$ 是 $1\times (m+n)$ 阵, 且$$\begin{array}{r}[XR{]}_{1,j}=\sum _{\ell =1}^{m+n}[X{]}_{1,\ell }[R{]}_{\ell ,j}=\sum _{1⩽\ell ⩽m+n}[R{]}_{\ell ,j}{x}^{m+n-\ell }.\end{array}$$若 $j⩽m$, 则 $[R{]}_{\ell ,j}=a_{\ell -j}$, 且$$\begin{array}{rl}[XR{]}_{1,j}=& \sum _{1⩽\ell ⩽m+n}{a}_{\ell -j}{x}^{m+n-\ell }\\ =& \sum _{1-j⩽\ell -j⩽m+n-j}{a}_{\ell -j}{x}^{n-(\ell -j)+(m-j)}\\ =& \sum _{1-j⩽h⩽n+(m-j)}{a}_h{x}^{n-h}{x}^{m-j}\\ =& \sum _{0⩽h⩽n}{a}_h{x}^{n-h}{x}^{m-j}+\sum _{\begin{array}{c}1-j⩽h<0\\ \text{或}n<h⩽n+(m-j)\end{array}}{a}_h{x}^{n-h}{x}^{m-j}\\ =& \left(\sum _{0⩽h⩽n}{a}_h{x}^{n-h}\right)x^{m-j}+\sum _{\begin{array}{c}1-j⩽h<0\\ \text{或}n<h⩽n+(m-j)\end{array}}0x^{n-h}{x}^{m-j}\\ =& f(x)x^{m-j}.\end{array}$$若 $j>m$, 则 $[R{]}_{\ell ,j}=b_{\ell -(j-m)}$. 为方便, 记 $k=j-m$. 则$$\begin{array}{rl}[XR{]}_{1,j}=& \sum _{1⩽\ell ⩽m+n}{b}_{\ell -k}{x}^{m+n-\ell }\\ =& \sum _{1-k⩽\ell -k⩽m+n-k}{b}_{\ell -k}{x}^{m-(\ell -k)+(n-k)}\\ =& \sum _{1-k⩽h⩽m+(n-k)}{b}_h{x}^{m-h}{x}^{n-k}\\ =& \sum _{0⩽h⩽m}{b}_h{x}^{m-h}{x}^{n-k}+\sum _{\begin{array}{c}1-k⩽h<0\\ \text{或}m<h⩽m+(n-k)\end{array}}{b}_h{x}^{m-h}{x}^{n-k}\\ =& \left(\sum _{0⩽h⩽m}{b}_h{x}^{m-h}\right)x^{n-(j-m)}+\sum _{\begin{array}{c}1-k⩽h<0\\ \text{或}m<h⩽m+(n-k)\end{array}}0x^{m-h}{x}^{n-k}\\ =& g(x)x^{n-(j-m)}.\end{array}$$

(3) 既然 $XR$ 是 $1\times (m+n)$ 阵, 则 $(XR)p$ 是 $1$ 级阵. 记 $r=[(XR)p{]}_{1,1}$. 则$$\begin{array}{rl}r=& \sum _{j=1}^{m+n}[XR{]}_{1,j}[p{]}_{j,1}\\ =& \sum _{j=1}^m[XR{]}_{1,j}[p{]}_{j,1}+\sum _{j=m+1}^{m+n}[XR{]}_{1,j}[p{]}_{j,1}\\ =& \sum _{j=1}^{m}f(x)x^{m-j}[p{]}_{j,1}+\sum _{j=m+1}^{m+n}g(x)x^{n-(j-m)}[p{]}_{j,1}\\ =& f(x)\sum _{j=1}^m{x}^{m-j}[p{]}_{j,1}+g(x)\sum _{j=m+1}^{m+n}{x}^{n-(j-m)}[p{]}_{j,1}\\ =& f(x)\sum _{k=0}^{m-1}[p{]}_{k+1,1}{x}^{m-1-k}+g(x)\sum _{k=0}^{n-1}[p{]}_{m+1+k,1}{x}^{n-1-k}.\end{array}$$记$$\begin{array}{rl}u(x)=\sum _{k=0}^{m-1}[p{]}_{k+1,1}{x}^{m-1-k}=& [p{]}_{1,1}{x}^{m-1}+[p{]}_{2,1}{x}^{m-2}+\cdots +[p{]}_{m,1},\\ v(x)=\sum _{k=0}^{n-1}[p{]}_{m+1+k,1}{x}^{n-1-k}=& [p{]}_{m+1,1}{x}^{n-1}+[p{]}_{m+2,1}{x}^{n-2}+\cdots +[p{]}_{m+n,1}.\end{array}$$则 $f(x)u(x)+g(x)v(x)=r$. 另一方面, $$\begin{array}{rl}r=& [(XR)p{]}_{1,1}=[X(Rp){]}_{1,1}\\ =& [X(\det (R)e){]}_{1,1}\\ =& \sum _{j=1}^{m+n}[X{]}_{1,j}[\det (R)e{]}_{j,1}\\ =& [X{]}_{1,m+n}[\det (R)e{]}_{m+n,1}\\ =& 1\det (R)=\det (R).\end{array}$$故 $f(x)u(x)+g(x)v(x)=\det (R)$.

(4) 设数 $c$ 适合 $f(c)=0=g(c)$. 由 (3) 知, 必 $0=f(c)u(c)+g(c)v(c)=\det (R)$ (我们代 $x$ 以数 $c$). 这有时是有用的.

例 78.7. 解方程组$$\{\begin{array}{l}11x^2-2xy-44y^2-x+26y-32=0,\\ 4x^2-18xy+49y^2-4x+9y-118=0.\end{array}$$(78.1)

我们可用上例的结果解方程组 (78.1). 为此, 我们记$$\begin{array}{rl}{f}_y(x)& =11x^2+(-2y-1)x+(-44y^2+26y-32),\\ g_y(x)& =4x^2+(-18y-4)x+(49y^2+9y-118).\end{array}$$设 $x=a$, $y=b$ 是方程组 (78.1) 的一个解. 则$$\begin{array}{rl}& \begin{array}{rl}0=& 11a^2-2ab-44b^2-a+26b-32\\ =& 11a^2+(-2b-1)a+(-44b^2+26b-32),\end{array}\\ & \begin{array}{rl}0=& 4a^2-18ab+49b^2-4a+9b-118\\ =& 4a^2+(-18b-4)a+(49b^2+9b-118),\end{array}\end{array}$$即 $f_b(a)=0=g_b(a)$. 故$$\begin{array}{r}\left[\begin{array}{cccc}11& 0& 4& 0\\ -2b-1& 11& -18b-4& 4\\ -44b^2+26b-32& -2b-1& 49b^2+9b-118& -18b-4\\ 0& -44b^2+26b-32& 0& 49b^2+9b-118\end{array}\right]\end{array}$$的行列式是 $0$. 另一方面, 可以算出, 此阵的行列式是 $342125(b^2-1)(b^2-4)$. 于是, 若 $x=a$, $y=b$ 是方程组 (78.1) 的一个解, 则 $b=1$ 或 $b=-1$ 或 $b=2$ 或 $b=-2$. 则 (代 $b$ 以 $1$) $$\{\begin{array}{l}11a^2-2a-44-a+26-32=0,\\ 4a^2-18a+49-4a+9-118=0;\end{array}$$(78.2)或 (代 $b$ 以 $-1$) $$\{\begin{array}{l}11a^2+2a-44-a-26-32=0,\\ 4a^2+18a+49-4a-9-118=0;\end{array}$$(78.3)或 (代 $b$ 以 $2$) $$\{\begin{array}{l}11a^2-4a-176-a+52-32=0,\\ 4a^2-36a+196-4a+18-118=0;\end{array}$$(78.4)或 (代 $b$ 以 $-2$) $$\{\begin{array}{l}11a^2+4a-176-a-52-32=0,\\ 4a^2+36a+196-4a-18-118=0.\end{array}$$(78.5)方程组 (78.2) 的解是 $a=-2$; 方程组 (78.3) 的解是 $a=3$; 方程组 (78.4) 的解是 $a=4$; 方程组 (78.5) 的解是 $a=-5$. 于是, 若 $x=a$, $y=b$ 是方程组 (78.1) 的一个解, 必: $$\begin{array}{rl}& a=-2,b=1;\\ \text{或}& a=3,b=-1;\\ \text{或}& a=4,b=2;\\ \text{或}& a=-5,b=-2.\end{array}$$最后, 我们验证这些是否是方程组 (78.1) 的解. 可以验证, 它们都是解. 于是, 方程组 (78.1) 的解是: $$\begin{array}{rl}& x=-2,y=1;\\ \text{或}& x=3,y=-1;\\ \text{或}& x=4,y=2;\\ \text{或}& x=-5,y=-2.\end{array}$$