高代 | 行列式 | 升阶法 | 拆分法与递推法

(天津大学,2021; 兰州大学,2021;南昌大学,2021; 南方科技大学,2021; 华南师范大学,2021; 东北大学,2021; 上海财经大学,2021; 西南财经大学,2020; 安徽大学,2020; 西北大学,2020; 华中科技大学,2020)计算行列式
${ D_{n}=\left|\begin{array}{ccccc} a_{1}+x_{1} & a_{2} & a_{3} & \cdots & a_{n} \\ a_{1} & a_{2}+x_{2} & a_{3} & \cdots & a_{n} \\ a_{1} & a_{2} & a_{3}+x_{3} & \cdots & a_{n} \\ \vdots & \vdots & \vdots & & \vdots \\ a_{1} & a_{2} & a_{3} & \cdots & a_{n}+x_{n} \end{array}\right| . }$

solution
利用升阶法,有
${ D_{n}=\left|\begin{array}{cccccc} 1 & a_{1} & a_{2} & a_{3} & \cdots & a_{n} \\ 0 & a_{1}+x_{1} & a_{2} & a_{3} & \cdots & a_{n} \\ 0 & a_{1} & a_{2}+x_{2} & a_{3} & \cdots & a_{n} \\ 0 & a_{1} & a_{2} & a_{3}+x_{3} & \cdots & a_{n} \\ \vdots & \vdots & \vdots & \vdots & & \vdots \\ 0 & a_{1} & a_{2} & a_{3} & \cdots & a_{n}+x_{n} \end{array}\right|=\left|\begin{array}{cccccc} 1 & a_{1} & a_{2} & a_{3} & \cdots & a_{n} \\ -1 & x_{1} & 0 & 0 & \cdots & 0 \\ -1 & 0 & x_{2} & 0 & \cdots & 0 \\ -1 & 0 & 0 & x_{3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots \\ -1 & 0 & 0 & 0 & \cdots & x_{n} \end{array}\right| . }$
当 ${x_{1} x_{2} \cdots x_{n} \neq 0}$ 时,依次将上述行列式的第 ${i(i=2,3,\cdots,n+1)}$ 列的 ${\dfrac{1}{x_{i}}}$ 倍加到第一列,就有
${ D_{n}=\left|\begin{array}{cccccc} 1+\sum\limits_{i=1}^{n} \dfrac{a_{i}}{x_{i}} & a_{1} & a_{2} & a_{3} & \cdots & a_{n} \\ 0 & x_{1} & 0 & 0 & \cdots & 0 \\ 0 & 0 & x_{2} & 0 & \cdots & 0 \\ 0 & 0 & 0 & x_{3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & 0 & \cdots & x_{n} \end{array}\right|=\left(1+\sum_{i=1}^{n} \frac{a_{i}}{x_{i}}\right) \prod_{j=1}^{n} x_{j}=\prod_{j=1}^{n} x_{j}+\sum_{i=1}^{n}\left(a_{i} \prod_{j \neq i} x_{j}\right) . }$
另外,根据行列式的运算规则可知 ${D_{n}}$ 是关于 ${x_{1},x_{2},\cdots,x_{n}}$ 的多项式,而上式右端也为 ${x_{1},x_{2},\cdots,x_{n}}$ 的多项式,它们在 ${x_{1} x_{2} \cdots x_{n} \neq 0}$ 的时候相等,自然在 ${x_{1} x_{2} \cdots x_{n}=0}$ 的时候也相等.即对任意的 ${x_{1},x_{2},\cdots,x_{n}}$ ,均有
${ D_{n}=\prod_{j=1}^{n} x_{j}+\sum_{i=1}^{n}\left(a_{i} \prod_{j \neq i} x_{j}\right) .}$

(华东理工大学,2021)求行列式
$D_{n}=\left|\begin{array}{ccccc} 1 & 2 & \cdots & n-1 & n+x \\ 1 & 2 & \cdots & n-1+x & n \\ \vdots & \vdots & & \vdots & \vdots \\ 1 & 2+x & \cdots & n-1 & n \\ 1+x & 2 & \cdots & n-1 & n \end{array}\right| .$

solution
当 ${x=0}$ 时,显然 ${D_{1}=1}$ ,且 ${D_{n}=0(n \geqslant 2)}$ .当 ${x \neq 0}$ 时,利用升阶法,有
$D_{n}=\left|\begin{array}{cccccc} 1 & 1 & 2 & \cdots & n-1 & n \\ 0 & 1 & 2 & \cdots & n-1 & n+x \\ 0 & 1 & 2 & \cdots & n-1+x & n \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 1 & 2+x & \cdots & n-1 & n \\ 0 & 1+x & 2 & \cdots & n-1 & n \end{array}\right|=\left|\begin{array}{cccccc} 1 & 1 & 2 & \cdots & n-1 & n \\ -1 & 0 & 0 & \cdots & 0 & x \\ -1 & 0 & 0 & \cdots & x & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ -1 & 0 & x & \cdots & 0 & 0 \\ -1 & x & 0 & \cdots & 0 & 0 \end{array}\right| .$

将上述行列式的第 ${i(i=2,3,\cdots,n+1)}$ 列的 ${\frac{1}{x}}$ 倍加到第一列,就有
$D_{n}=\left|\begin{array}{cccccc} 1+\dfrac{n(n+1)}{2 x} & 1 & 2 & \cdots & n-1 & n \\ 0 & 0 & 0 & \cdots & 0 & x \\ 0 & 0 & 0 & \cdots & x & 0 \\ \vdots& \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & x & \cdots & 0 & 0 \\ 0 & x & 0 & \cdots & 0 & 0 \end{array}\right|=(-1)^{\frac{n(n-1)}{2}}\left[x+\frac{n(n+1)}{2}\right] x^{n-1}$

明显上式右端的结果对 ${x=0}$ 也成立.

拆分法与递推法

(北京工业大学,2021; 山东师范大学,2021)计算 $n$ 阶行列式
$D_{n}=\left|\begin{array}{cccccc} x & a & a & \cdots & a & a \\ -a & x & a & \cdots & a & a \\ -a & -a & x & \cdots & a & a \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ -a & -a & -a & \cdots & x & a \\ -a & -a & -a & \cdots & -a & x \end{array}\right|$

solution
当 $a=0$ 时,显然有 $D_{n}=x^{n}$ .当 $a \neq 0$ 时,将 $D_{n}$ 按照第一列拆为两个行列式,有
$\begin{aligned} D_{n} & =\left|\begin{array}{cccccc} x+a & a & a & \cdots & a & a \\ 0 & x & a & \cdots & a & a \\ 0 & -a & x & \cdots & a & a \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & -a & -a & \cdots & x & a \\ 0 & -a & -a & \cdots & -a & x \end{array}\right|+\left|\begin{array}{cccccc} -a & a & a & \cdots & a & a \\ -a & x & a & \cdots & a & a \\ -a & -a & x & \cdots & a & a \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ -a & -a & -a & \cdots & x & a \\ -a & -a & -a & \cdots & -a & x \end{array}\right| \\ & =(x+a) D_{n-1}+\left|\begin{array}{ccccccc} -a & 0 & 0 & \cdots & 0 & 0 & \\ -a & x-a & 0 & \cdots & 0 & 0 \\ -a & -2 a & x-a & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ -a & -2 a & -2 a & \cdots & x-a & 0 \\ -a & -2 a & -2 a & \cdots & -2 a & x-a \end{array}\right| \\ & =(x+a) D_{n-1}-a(x-a)^{n-1} . \end{aligned}$
那么根据对称性,还有 $D_{n}=(x-a) D_{n-1}+a(x+a)^{n-1}$ ,进而
$\left\{\begin{array}{l} (x-a) D_{n}=(x-a)(x+a) D_{n-1}-a(x-a)^{n} \\ (x+a) D_{n}=(x-a)(x+a) D_{n-1}+a(x+a)^{n} \end{array}\right.$

上述两式相减可得
$D_{n}=\frac{(x+a)^{n}+(x-a)^{n}}{2} .$
显然上式对 $a=0$ 也成立.

(华南师范大学,2020)计算 $n$ 阶行列式
$D_{n}=\left|\begin{array}{ccccc} \sqrt{5} & 1 & & & \\ 1 & \sqrt{5} & 1 & & \\ & 1 & \ddots & \ddots & \\ & & \ddots & \sqrt{5} & 1 \\ & & & 1 & \sqrt{5} \end{array}\right| .$

solution
将 $D_{n}$ 按照第一列展开,可知
$D_{n}=\sqrt{5} D_{n-1}-D_{n-2} .\quad (1)$
设方程 $x^{2}-\sqrt{5} x+1=0$ 的两个根为
$\alpha=\frac{\sqrt{5}-1}{2},\beta=\frac{\sqrt{5}+1}{2} .$
则由韦达定理可知 $\alpha+\beta=\sqrt{5},\alpha \beta=1$ ,所以(1)式等价于
$D_{n}=(\alpha+\beta) D_{n-1}-\alpha \beta D_{n-2} .$
变形就有
$D _ { n } - \alpha D _ { n - 1 } = \beta ( D _ { n - 1 } - \alpha D _ { n - 2 } ) ;\quad (2)$
$D _ { n } - \beta D _ { n - 1 } = \alpha ( D _ { n - 1 } - \beta D _ { n - 2 } ) .\quad (3)$
注意到
$D_{1}=\sqrt{5}=\alpha+\beta$

$\begin{aligned}D_{2}&=\left|\begin{array}{cc}\sqrt{5} & 1 \\ 1 & \sqrt{5}\end{array}\right|\\&=5-1\\&=(\alpha+\beta)^{2}-\alpha \beta\\&=\alpha^{2}+\alpha \beta+\beta^{2}\end{aligned}$
于是
$D_{2}-\alpha D_{1}=\beta^{2},D_{2}-\beta D_{1}=\alpha^{2} .$
从而结合(2)(3)式可知
$D_{n}-\alpha D_{n-1}=\beta\left(D_{n-1}-\alpha D_{n-2}\right)=\cdots=\beta^{n-2}\left(D_{2}-\alpha D_{1}\right)=\beta^{n} \quad (4)$
$D_{n}-\beta D_{n-1}=\alpha\left(D_{n-1}-\beta D_{n-2}\right)=\cdots=\alpha^{n-2}\left(D_{2}-\beta D_{1}\right)=\alpha^{n} \quad (5)$
那么(4)式乘以 $\beta$ 与(5)式乘以 $\alpha$ 相减可得
$D_{n}=\frac{\beta^{n+1}-\alpha^{n+1}}{\beta-\alpha}=\left(\frac{\sqrt{5}+1}{2}\right)^{n+1}-\left(\frac{\sqrt{5}-1}{2}\right)^{n+1} .$

高代 | 行列式 | 升阶法 | 拆分法与递推法

拆分法与递推法

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