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习题一26(3)(5):
计算下列$n$阶行列式:
(3)
$$\left| {\begin{array}{*{20}{c}}
{a_1b_1}&{a_1b_2}&{a_1b_3}&{\cdots}&{a_1b_n}\\
{a_1b_2}&{a_2b_2}&{a_2b_3}&{\cdots}&{a_2b_n}\\
{a_1b_3}&{a_2b_3}&{a_3b_3}&{\cdots}&{a_3b_n}\\
{\vdots}&{\vdots}&{\vdots}&{}&{\vdots}\\
{a_1b_n}&{a_2b_n}&{a_3b_n}&{\cdots}&{a_nb_n}
\end{array}} \right|$$
(5)
$$\left| {\begin{array}{*{20}{c}}
{2\cos\alpha}&{1}&{0}&{\cdots}&{\cdots}&{0}\\
{1}&{2\cos\alpha}&{1}&{\ddots}&{}&{\vdots}\\
{0}&{1}&{2\cos\alpha}&{\ddots}&{\ddots}&{\vdots}\\
{\vdots}&{\ddots}&{\ddots}&{\ddots}&{\ddots}&{0}\\
{\vdots}&{}&{\ddots}&{\ddots}&{\ddots}&{1}\\
{0}&{\cdots}&{\cdots}&{0}&{1}&{2\cos\alpha}
\end{array}} \right|$$
解:
(3)此题有两个解法
解法1:
记
$$D_n=\left| {\begin{array}{*{20}{c}}
{a_1b_1}&{a_1b_2}&{a_1b_3}&{\cdots}&{a_1b_n}\\
{a_1b_2}&{a_2b_2}&{a_2b_3}&{\cdots}&{a_2b_n}\\
{a_1b_3}&{a_2b_3}&{a_3b_3}&{\cdots}&{a_3b_n}\\
{\vdots}&{\vdots}&{\vdots}&{}&{\vdots}\\
{a_1b_n}&{a_2b_n}&{a_3b_n}&{\cdots}&{a_nb_n}
\end{array}} \right|$$
当
$$b_i \ne 0,i=1,2,\cdots,n-1$$
时
$$\begin{eqnarray*}D_n&=&\left| {\begin{array}{*{20}{c}}
{a_1b_1}&{a_1b_2}&{a_1b_3}&{\cdots}&{a_1b_n}\\
{a_1b_2}&{a_2b_2}&{a_2b_3}&{\cdots}&{a_2b_n}\\
{a_1b_3}&{a_2b_3}&{a_3b_3}&{\cdots}&{a_3b_n}\\
{\vdots}&{\vdots}&{\vdots}&{}&{\vdots}\\
{a_1b_n}&{a_2b_n}&{a_3b_n}&{\cdots}&{\frac{a_{n-1}b_n^2}{b_{n-1}}}
\end{array}} \right|+
\left| {\begin{array}{*{20}{c}}
{a_1b_1}&{a_1b_2}&{a_1b_3}&{\cdots}&{a_1b_n}\\
{a_1b_2}&{a_2b_2}&{a_2b_3}&{\cdots}&{a_2b_n}\\
{a_1b_3}&{a_2b_3}&{a_3b_3}&{\cdots}&{a_3b_n}\\
{\vdots}&{\vdots}&{\vdots}&{}&{\vdots}\\
{0}&{0}&{0}&{\cdots}&{a_nb_n-\frac{a_{n-1}b_n^2}{b_{n-1}}}
\end{array}} \right|\\
&=&\frac{b_n}{b_{n-1}}(a_nb_{n-1}-a_{n-1}b_n)D_{n-1}\\
&=&\cdots\\
&=&(-1)^{n+1}a_1b_n\prod\limits_{i=1}^{n-1}(a_ib_{i+1}-a_{i+1}b_i)
\end{eqnarray*}$$
当$b_i$至少有一个为$0$
$$i=1,2,\cdots,n-1$$
时
可以归纳假设上式对
$$n=k-2$$
时成立,当
$$n=k-1$$
时,若
$$b_{n-1} \ne 0$$
则已经成立
若
$$b_{n-1}=0$$
由于行列式展开后是多项式函数,多项式是连续函数
可以用可去间断点的极限值来代替未定义的函数值,进行连续延拓,则也成立
或者,可以固定其他元,将$n=k-1$时的多元多项式视为关于$b_{n-1}$的一元多项式
由于,此多项式和上式在无穷多个点上取值相同,所以这两个多项式相等
解法2:
记
$$D_n=\left| {\begin{array}{*{20}{c}}
{a_1b_1}&{a_1b_2}&{a_1b_3}&{\cdots}&{a_1b_n}\\
{a_1b_2}&{a_2b_2}&{a_2b_3}&{\cdots}&{a_2b_n}\\
{a_1b_3}&{a_2b_3}&{a_3b_3}&{\cdots}&{a_3b_n}\\
{\vdots}&{\vdots}&{\vdots}&{}&{\vdots}\\
{a_1b_n}&{a_2b_n}&{a_3b_n}&{\cdots}&{a_nb_n}
\end{array}} \right|$$
于是
$$\begin{eqnarray*}D_n&=&a_1\left| {\begin{array}{*{20}{c}}
{b_1}&{a_1b_2}&{a_1b_3}&{\cdots}&{a_1b_n}\\
{b_2}&{a_2b_2}&{a_2b_3}&{\cdots}&{a_2b_n}\\
{b_3}&{a_2b_3}&{a_3b_3}&{\cdots}&{a_3b_n}\\
{\vdots}&{\vdots}&{\vdots}&{}&{\vdots}\\
{b_n}&{a_2b_n}&{a_3b_n}&{\cdots}&{a_nb_n}
\end{array}} \right|\\
&=&a_1\left| {\begin{array}{*{20}{c}}
{b_1}&{a_1b_2-a_2b_1}&{a_1b_3-a_3b_1}&{\cdots}&{a_1b_n-a_nb_1}\\
{b_2}&{0}&{a_2b_3-a_3b_2}&{\cdots}&{a_2b_n-a_nb_2}\\
{b_3}&{0}&{0}&{\cdots}&{a_3b_n-a_nb_3}\\
{\vdots}&{\vdots}&{\vdots}&{}&{\vdots}\\
{b_n}&{0}&{0}&{\cdots}&{0}
\end{array}} \right|\\
&=&(-1)^{n-1}a_1\left| {\begin{array}{*{20}{c}}
{a_1b_2-a_2b_1}&{a_1b_3-a_3b_1}&{\cdots}&{a_1b_n-a_nb_1}&{b_1}\\
{0}&{a_2b_3-a_3b_2}&{\cdots}&{a_2b_n-a_nb_2}&{b_2}\\
{0}&{0}&{\cdots}&{a_3b_n-a_nb_3}&{b_3}\\
{\vdots}&{\vdots}&{}&{\vdots}&{\vdots}\\
{0}&{0}&{\cdots}&{0}&{b_n}
\end{array}} \right|\\
&=&(-1)^{n+1}a_1b_n\prod\limits_{i=1}^{n-1}(a_ib_{i+1}-a_{i+1}b_i)
\end{eqnarray*}$$
(5)
记
$$D_n=\left| {\begin{array}{*{20}{c}}
{2\cos\alpha}&{1}&{0}&{\cdots}&{\cdots}&{0}\\
{1}&{2\cos\alpha}&{1}&{\ddots}&{}&{\vdots}\\
{0}&{1}&{2\cos\alpha}&{\ddots}&{\ddots}&{\vdots}\\
{\vdots}&{\ddots}&{\ddots}&{\ddots}&{\ddots}&{0}\\
{\vdots}&{}&{\ddots}&{\ddots}&{\ddots}&{1}\\
{0}&{\cdots}&{\cdots}&{0}&{1}&{2\cos\alpha}
\end{array}} \right|$$
按第一(行)列展开
$$D_n=2\cos\alpha D_{n-1}-D_{n-2}$$
也即
$$D_n-2\cos\alpha D_{n-1}+D_{n-2}=0$$
此差分方程的特征方程是
$$\lambda^2-2\cos\alpha\lambda+1=0$$
解得一对共轭复根
$$\lambda_{1,2}=\cos\alpha \pm i\sin\alpha$$
根据$De Moivre$公式
$$\begin{eqnarray*}
D_n&=&A(\cos (n-1)\alpha+i\sin (n-1)\alpha)+B(\cos (n-1)\alpha-i\sin (n-1)\alpha)\\
&=&(A+B)\cos (n-1)\alpha+(A-B)i\sin (n-1)\alpha
\end{eqnarray*}$$
由于$D_1=2\cos\alpha$,$D_2=4\cos^2\alpha-1$,代入可解得
$$\left\{ \begin{array}{l}
A+B=2\cos\alpha\\
(A-B)i=\frac{\cos 2\alpha}{\sin\alpha}
\end{array} \right.$$
所以,当$\sin\alpha \ne 0$,也即$\alpha \ne k\pi$时
$$\begin{eqnarray*}
D_n&=&2\cos\alpha\cos (n-1)\alpha+\frac{\cos 2\alpha}{\sin\alpha}\sin (n-1)\alpha\\
&=&\frac{\sin 2\alpha\cos(n-1)\alpha+\cos 2\alpha\sin(n-1)\alpha}{\sin\alpha}\\
&=&\frac{\sin (n+1)\alpha}{\sin\alpha}
\end{eqnarray*}$$
当
$$\alpha = k\pi$$
时
由于行列式展开后是多项式函数,它是连续函数,三角函数也是连续函数
所以它们的复合函数也是连续函数,可以用可去间断点的极限值来代替未定义的函数值,进行连续延拓
$$\begin{eqnarray*}
D_n&=&\lim\limits_{\alpha \to k\pi}\frac{\sin (n+1)\alpha}{\sin\alpha}\\
&=&\lim\limits_{\alpha \to k\pi}(n+1)\cos n\alpha(L'Hospital法则)\\
&=&(n+1)\cos^n \alpha
\end{eqnarray*}$$
综上所述
$$D_n=\left\{ \begin{array}{l}
\frac{\sin (n+1)\alpha}{\sin\alpha}, \alpha \ne k\pi\\
(n+1)\cos^n \alpha, \alpha = k\pi
\end{array} \right.$$
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