标签：展开 假设 n+1 ·· rod mat 矩阵 span 进一步
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\(Page.101\\1.\ D(A)=D(A^{(1)},A^{(2)},A^{(3)},A^{(4)})=D(1000A^{(1)}+100A^{(2)}+10A^{(3)}+A^{(4)},A^{(2)},A^{(3)},A^{(4)})\\\quad 又\ 1000A^{(1)}+100A^{(2)}+10A^{(3)}+A^{(4)}=\begin{pmatrix}1798\\2139\\3255\\4867\end{pmatrix}\ 内元素均可被31整除\\\quad 依第一列展开可知\ D(A)=D(A^{(1)},A^{(2)},A^{(3)},A^{(4)})\ 可被31整除\)?
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\(2.\ 设A为任意n阶斜对称行列式,则必存在n阶可逆矩阵P,使得\ A=P^{T}BP\ ,且
\\\quad detB=\left\{\begin{array}{l}1\quad n\in 2k\\0\quad n\in 2k+1\end{array}\right.
\\\quad detA=det^{2}P*detB=\left\{\begin{array}{l}det^{2}P\quad n\in 2k\\0\quad n\in 2k+1\end{array}\right.\)
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\(3.\ 略\)
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\(4.\ 讨论消除第一个 M_{k_{1}}^{n}(x_{1}) 块时(假设k_{1}=max(k_{i})),其余M_{k_{i}}^{n}(x_{i})块的变化\\\quad
显然在初始状态下M_{k_{i}}^{n}(x_{i})(a,b)=\left\{\begin{array}{l}\binom{b-1}{a-1}x_{i}^{b-2}\qquad a>b\\0\qquad else.\end{array}\right. \quad 下面的讨论中我们仅关注a>b的情况\\\quad
定义变换(a):A^{(i)}=A^{(i)}-x_{1}*A^{(i-1)};变换(b):A^{(i)}=A^{(i)}-x_{i}*A^{(i-1)}\\\quad
那么在第一次变换(a)中,M_{k_{i}}^{n}(x_{i})(a,b)=M_{k_{i}}^{n}(x_{i})(a,b)-x_{1}*M_{k_{i}}^{n}(x_{i})(a,b-1)=\binom{b-1}{a-1}x_{i}^{b-2}-\binom{b-1}{a-2}x_{i}^{b-3}\\\quad
以此类推,根据排列组合公式计算得出第k次变换(a)后M_{k_{i}}^{n}(x_{i})(a,b)=\sum_{j=0}^{k}\binom{k}{j}\binom{b-j-1}{a-1}(-1)^{j}x_{i}^{b-j-2}x_{1}^{j}\\\quad
总结得，每次变换(a)使得k=k+1;变换(b)使得b=b-1\\\quad
当b=0时,除第一行每行即可提取出(x_{j}-x_{1})^{k_{1}}，共提取出(x_{j}-x_{1})^{k_{1}(k_{j}-1)};第一行可提取出(x_{j}-x_{1})^{k_{1}}\\\quad
综上当m=2时,\Delta =(x_{1}-x_{2})^{k_{1}(k_{2}-1)+k_{1}}=(x_{1}-x_{2})^{k_{1}k_{2}}\\\quad
依次类推我们可以得到\Delta _{n}(k_{1},x_{1};···;k_{n},x_{n})=\prod_{1\leqslant j\leqslant i\leqslant m}(x_{i}-x_{j})^{k_{i}k_{j}}\)
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\(5.\ B _{n}=\begin{vmatrix}
1&s-t&\frac{(s-t)(s-t+1)}{2!}&···&\frac{(s-t)···(s-t+n-2)}{(n-1)!}\\
1&s-t+1&\frac{(s-t+1)(s-t+2)}{2!}&···&\frac{(s-t+1)···(s-t+n-1)}{(n-1)!}\\
1&s-t+2&\frac{(s-t+2)(s-t+3)}{2!}&···&\frac{(s-t+2)···(s-t+n)}{(n-1)!}\\
···&···&···&···&···\\
1&s-t+n&\frac{(s-t+n)(s-t+n+1)}{2!}&···&\frac{(s-t+n)···(s-t+2n-2)}{(n-1)!}\end{vmatrix}*\prod_{i=1}^{t}\frac{\binom{n+s-i}{n}}{\binom{n+t-i}{n}}\\\ \ \qquad
=\begin{vmatrix}
1&s-t&\frac{(s-t)(s-t+1)}{2!}&···&\frac{(s-t)···(s-t+n-2)}{(n-2)!}\\
0&1&s-t+1&···&\frac{(s-t+1)···(s-t+n-2)}{(n-2)!}\\
0&1&s-t+2&···&\frac{(s-t+2)···(s-t+n-1)}{(n-2)!}\\
···&···&···&···&···\\
0&1&s-t+n&···&\frac{(s-t+n)···(s-t+2n-3)}{(n-2)!}\end{vmatrix}*\prod_{i=1}^{t}\frac{\binom{n+s-i}{n}}{\binom{n+t-i}{n}}\\\ \ \qquad
=·········\\\ \ \qquad
=\begin{vmatrix}
1&s-t&\frac{(s-t)(s-t+1)}{2!}&···&\frac{(s-t)···(s-t+n-2)}{(n-1)!}\\
0&1&s-t+1&···&\frac{(s-t+1)···(s-t+n-2)}{(n-2)!}\\
0&0&1&···&\frac{(s-t+2)···(s-t+n-2)}{(n-3)!}\\
···&···&···&···&···\\
0&0&0&···&1\end{vmatrix}*\prod_{i=1}^{t}\frac{\binom{n+s-i}{n}}{\binom{n+t-i}{n}}\\\ \ \qquad
=\prod_{i=1}^{t}\frac{\ \binom{n+s-i}{n}\ }{\ \binom{n+t-i}{n}\ }\)
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\(6.\ C_{n}=\begin{vmatrix}
\lambda _{1}&1&&···&&\\
-1&\lambda _{2}&1&···&&\\
&-1&\lambda _{3}&···&&\\
···&···&···&···&···&···\\
&&&···&\lambda _{n-1}&1\\
&&&···&-1&\lambda _{n}
\end{vmatrix}\\\ \ \qquad
=\lambda _{n}C_{n-1}-\begin{vmatrix}
\lambda _{1}&1&&···&&\\
-1&\lambda _{2}&1&···&&\\
&-1&\lambda _{3}&···&&\\
···&···&···&···&···&···\\
&&&···&\lambda _{n-2}&1\\
&&&···&&-1
\end{vmatrix}\\\ \ \qquad
=\lambda _{n}C_{n-1}-\begin{vmatrix}
\lambda _{1}&1&&···&&\\
-1&\lambda _{2}&1&···&&\\
&-1&\lambda _{3}&···&&\\
···&···&···&···&···&···\\
&&&···&\lambda _{n-2}&0\\
&&&···&&-1
\end{vmatrix}\\\ \ \qquad
=\lambda _{n}C_{n-1}-C_{n-2}\)
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\(7.\ A_{n}=\begin{vmatrix}
2&-1& &···& & \\
-1&2&-1&···& & \\
&-1&2&···& & \\
···&···&···&···&···&···\\
& & &···&2&-1\\
& & &···&-1&2
\end{vmatrix}\\\ \ \qquad
=\begin{vmatrix}
1&-1& &···& & \\
0&2&-1&···& & \\
0&-1&2&···& & \\
···&···&···&···&···&···\\
0& & &···&2&-1\\
1& & &···&-1&2
\end{vmatrix}\\\ \ \qquad
=A_{n-1}+(-1)^{i+2}*(-1)^{i}\\\ \ \qquad
=A_{n-1}+1\\\ \ \qquad
=A_{1}+n-1\\\ \ \qquad
=n+1\)
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\(8.\ 利用分块矩阵，略\)
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\(9.\ 略\)
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标签：展开 假设 n+1 ·· rod mat 矩阵 span 进一步
原文地址：https://www.cnblogs.com/cihua/p/Dsxyl_1_3_2.html
