高代 | 行列式 | 加边法与分解法 | 拆分法与递推法

行列式

加边法与分解法

(中山大学,2022)计算行列式
$D=\left|\begin{array}{ccccc} 2 & 1 & 1 & 1 & 1 \\ 1 & \dfrac{3}{2} & 1 & 1 & 1 \\ 1 & 1 & \dfrac{4}{3} & 1 & 1 \\ 1 & 1 & 1 & \dfrac{5}{4} & 1 \\ 1 & 1 & 1 & 1 & \dfrac{6}{5} \end{array}\right|.$

solution
首先利用加边法将 ${D}$ 转化为 6 阶行列式,然后用第一行的负一倍加到之后的每一行,最后再将第 ${i(2 \leqslant i \leqslant 6)}$ 列的 ${i-1}$ 倍均加到第一列,有
$D=\left|\begin{array}{llllll} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 2 & 1 & 1 & 1 & 1 \\ 0 & 1 & \dfrac{3}{2} & 1 & 1 & 1 \\ 0 & 1 & 1 & \dfrac{4}{3} & 1 & 1 \\ 0 & 1 & 1 & 1 & \dfrac{5}{4} & 1 \\ 0 & 1 & 1 & 1 & 1 & \dfrac{6}{5} \end{array}\right|=\left|\begin{array}{cccccc} 1 & 1 & 1 & 1 & 1 & 1 \\ -1 & 1 & 0 & 0 & 0 & 0 \\ -1 & 0 & \dfrac{1}{2} & 0 & 0 & 0 \\ -1 & 0 & 0 & \dfrac{1}{3} & 0 & 0 \\ -1 & 0 & 0 & 0 & \dfrac{1}{4} & 0 \\ -1 & 0 & 0 & 0 & 0 & \dfrac{1}{5} \end{array}\right|=\left|\begin{array}{cccccc} 16 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \dfrac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & \dfrac{1}{3} & 0 & 0 \\ 0 & 0 & 0 & 0 & \dfrac{1}{4} & 0 \\ 0 & 0 & 0 & 0 & 0 & \dfrac{1}{5} \end{array}\right|=\dfrac{2}{15} .$

(东北师范大学,2022)计算行列式
$D_{n}=\left|\begin{array}{cccc} a_{1}+b_{1} & a_{1}+b_{2} & \cdots & a_{1}+b_{n} \\ a_{2}+b_{1} & a_{2}+b_{2} & \cdots & a_{2}+b_{n} \\ \vdots & \vdots & & \vdots \\ a_{n}+b_{1} & a_{n}+b_{2} & \cdots & a_{n}+b_{n} \end{array}\right|$

solution
首先容易发现
$D_{1}=a_{1}+b_{1}$
$\begin{aligned}D_{2}&=\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right)-\left(a_{1}+b_{2}\right)\left(a_{2}+b_{1}\right)\\&=\left(a_{1}-a_{2}\right)\left(b_{2}-b_{1}\right) .\end{aligned}$
另外,由于
$\begin{aligned}A & \triangleq\left(\begin{array}{cccc} a_{1}+b_{1} & a_{1}+b_{2} & \cdots & a_{1}+b_{n} \\ a_{2}+b_{1} & a_{2}+b_{2} & \cdots & a_{2}+b_{n} \\ \vdots & \vdots & & \vdots \\ a_{n}+b_{1} & a_{n}+b_{2} & \cdots & a_{n}+b_{n} \end{array}\right)\\&=\left(\begin{array}{cc} a_{1} & 1 \\ a_{2} & 1 \\ \vdots & \vdots \\ a_{n} & 1 \end{array}\right)\left(\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ b_{1} & b_{2} & \cdots & b_{n} \end{array}\right)\end{aligned}$
由此可知 ${r(A) \leqslant 2}$ ,所以当 ${n \geqslant 3}$ 时,有 ${D_{n}=|A|=0}$ .

(南京师范大学,2022)计算行列式
$D_{n}=\left|\begin{array}{cccc} 1+x_{1}^{2} & x_{1} x_{2} & \cdots & x_{1} x_{n} \\ x_{2} x_{1} & 1+x_{2}^{2} & \cdots & x_{2} x_{n} \\ \vdots & \vdots & & \vdots \\ x_{n} x_{1} & x_{n} x_{2} & \cdots & 1+x_{n}^{2} \end{array}\right| .$

solution
先将 ${D_{n}}$ 加边至 ${n+1}$ 阶,然后依次将第 1 行的 ${-x_{i}(1 \leqslant i \leqslant n)}$ 倍加到第 ${i+1}$ 行,再依次将第 ${i(2 \leqslant i \leqslant n+1)}$ 列的 ${x_{i-1}}$ 倍加到第 1 列,可得
$\begin{aligned} D_{n} &=\left|\begin{array}{ccccc} 1 & x_{1} & x_{2} & \cdots & x_{n} \\ 0 & 1+x_{1}^{2} & x_{1} x_{2} & \cdots & x_{1} x_{n} \\ 0 & x_{2} x_{1} & 1+x_{2}^{2} & \cdots & x_{2} x_{n} \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & x_{n} x_{1} & x_{n} x_{2} & \cdots & 1+x_{n}^{2} \end{array}\right|\\&=\left|\begin{array}{ccccc} 1 & x_{1} & x_{2} & \cdots & x_{n} \\ -x_{1} & 1 & 0 & \cdots & 0 \\ -x_{2} & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ -x_{n} & 0 & 0 & \cdots & 1 \end{array}\right|\\&=\left|\begin{array}{ccccc} 1+\sum\limits_{i=1}^{n} x_{i}^{2} & x_{1} & x_{2} & \cdots & x_{n} \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{array}\right|\\ &=1+\sum_{i=1}^{n} x_{i}^{2} . \end{aligned}$

(兰州大学,2022)计算行列式
$D_{n}=\left|\begin{array}{cccc} x_{1}+a_{1} b_{1} & x_{1}+a_{1} b_{2} & \cdots & x_{1}+a_{1} b_{n} \\ x_{2}+a_{2} b_{1} & x_{2}+a_{2} b_{2} & \cdots & x_{2}+a_{2} b_{n} \\ \vdots & \vdots & & \vdots \\ x_{n}+a_{n} b_{1} & x_{n}+a_{n} b_{2} & \cdots & x_{n}+a_{n} b_{n} \end{array}\right| .$

solution
明显 $D_{1}=x_{1}+a_{1} b_{1}$ ,同时
$\begin{aligned}D_{2}&=\left|\begin{array}{cc} x_{1}+a_{1} b_{1} & x_{1}+a_{1} b_{2} \\ x_{2}+a_{2} b_{1} & x_{2}+a_{2} b_{2} \end{array}\right|\\&=\left|\left(\begin{array}{cc} x_{1} & a_{1} \\ x_{2} & a_{2} \end{array}\right)\left(\begin{array}{cc} 1 & 1 \\ b_{1} & b_{2} \end{array}\right)\right|\\&=\left(x_{1} a_{2}-x_{2} a_{1}\right)\left(b_{2}-b_{1}\right) .\end{aligned}$
而当 ${n \geqslant 3}$ 时,由于
$\begin{aligned}A &\triangleq\left(\begin{array}{cccc} x_{1}+a_{1} b_{1} & x_{1}+a_{1} b_{2} & \cdots & x_{1}+a_{1} b_{n} \\ x_{2}+a_{2} b_{1} & x_{2}+a_{2} b_{2} & \cdots & x_{2}+a_{2} b_{n} \\ \vdots & \vdots & & \vdots \\ x_{n}+a_{n} b_{1} & x_{n}+a_{n} b_{2} & \cdots & x_{n}+a_{n} b_{n} \end{array}\right) \\ &= \left(\begin{array}{cc} x_{1} & a_{1} \\ x_{2} & a_{2} \\ \vdots & \vdots \\ x_{n} & a_{n} \end{array}\right)\left(\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ b_{1} & b_{2} & \cdots & b_{n} \end{array}\right)\end{aligned}$

由此可知 ${r(A) \leqslant 2}$ ,从而 ${A}$ 不可逆,即 $D_{n}=|A|=0$ .

拆分法与递推法

(陕西师范大学,2022; 兰州大学,2022)计算 ${n}$ 阶行列式
$D_{n}=\left|\begin{array}{ccccc} x & a & a & \cdots & a \\ -a & x & a & \cdots & a \\ -a & -a & x & \cdots & a \\ \vdots & \vdots & \vdots & & \vdots \\ -a & -a & -a & \cdots & x \end{array}\right| .$

solution
当 ${a=0}$ 时,显然 ${D_{n}=x^{n}}$ .当 ${a \neq 0}$ 时,由于
$\begin{aligned} D_{n}&=\left|\begin{array}{ccccc} (x+a)-a & a & a & \cdots & a \\ 0-a & x & a & \cdots & a \\ 0-a & -a & x & \cdots & a \\ \vdots & \vdots & \vdots & & \vdots \\ 0-a & -a & -a & \cdots & x \end{array}\right|\\&=\left|\begin{array}{ccccc} x+a & a & a & \cdots & a \\ 0 & x & a & \cdots & a \\ 0 & -a & x & \cdots & a \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & -a & -a & \cdots & x \end{array}\right|+\left|\begin{array}{ccccc} -a & a & a & \cdots & a \\ -a & x & a & \cdots & a \\ -a & -a & x & \cdots & a \\ \vdots & \vdots & \vdots & & \vdots \\ -a & -a & -a & \cdots & x \end{array}\right|\\ &=(x+a) D_{n-1}+\left|\begin{array}{ccccc} -a & 0 & 0 & \cdots & 0 \\ -a & x-a & 0 & \cdots & 0 \\ -a & -2 a & x-a & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ -a & -2 a & -2 a & \cdots & x-a \end{array}\right|\\&=(x+a) D_{n-1}-a(x-a)^{n-1} \end{aligned}$
利用对称性,也有 ${D_{n}=(x-a) D_{n-1}+a(x+a)^{n-1}}$ ,于是
${ \left\{\begin{array}{l} (x-a) D_{n}=(x+a)(x-a) D_{n-1}-a(x-a)^{n} \\ (x+a) D_{n}=(x+a)(x-a) D_{n-1}+a(x+a)^{n} \end{array}\right. }$
上述两式相减可得
$\begin{aligned} D_{n}&=\frac{a(x+a)^{n}+a(x-a)^{n}}{2 a}\\&=\frac{(x+a)^{n}+(x-a)^{n}}{2} .\end{aligned}$
明显上式右端对 ${a=0}$ 也成立.

(安徽大学,2022; 河北工业大学,2022)计算 ${n}$ 阶行列式
$D_{n}=\left|\begin{array}{cccccc} 2 a & a^{2} & & & & \\ 1 & 2 a & a^{2} & & & \\ & 1 & 2 a & a^{2} & & \\ & & \ddots & \ddots & \ddots & \\ & & & 1 & 2 a & a^{2} \\ & & & & 1 & 2 a \end{array}\right|$
其中 $a \neq 0.$

solution
将 ${D_{n}}$ 按照第一行展开,可得
${ D_{n}=2 a D_{n-1}-a^{2} D_{n-2} . }$
即 ${D_{n}-a D_{n-1}=a\left(D_{n-1}-a D_{n-2}\right)}$ ,递推可得
$\begin{aligned} D_{n}-a D_{n-1}&=a^{n-2}\left(D_{2}-a D_{1}\right)\\&=a^{n-2}\left(3 a^{2}-2 a^{2}\right)\\&=a^{n} .\end{aligned}$
于是
${ \frac{D_{n}}{a^{n}}-\frac{D_{n-1}}{a^{n-1}}=1 . }$
即 ${\left\{\dfrac{D_{n}}{a^{n}}\right\}}$ 是以 1 为公差的等差数列,进而
${ \frac{D_{n}}{a^{n}}=\frac{D_{1}}{a}+(n-1)=n+1 . }$
所以 ${D_{n}=(n+1) a^{n}}$ .

(东北大学,2022)计算 ${n}$ 阶行列式
$D_{n}=\left|\begin{array}{ccccc} a b & a & a & \cdots & a \\ b & a^{2} b^{2} & a & \cdots & a \\ b & b & a^{3} b^{3} & \cdots & a \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \cdots & a^{n} b^{n} \end{array}\right|$

solution
由于
$\begin{aligned} D_{n}&=\left|\begin{array}{cccccc} a b & a & a & \cdots & a & a \\ b & a^{2} b^{2} & a & \cdots & a & a \\ b & b & a^{3} b^{3} & \cdots & a & a \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ b & b & b & \cdots & a^{n-1} b^{n-1} & a \\ b & b & b & \cdots & b & a^{n} b^{n} \end{array}\right|\\&=\left|\begin{array}{cccccc} a b & a & a & \cdots & a & a+0 \\ b & a^{2} b^{2} & a & \cdots & a & a+0 \\ b & b & a^{3} b^{3} & \cdots & a & a+0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ b & b & b & \cdots & a^{n-1} b^{n-1} & a+0 \\ b & b & b & \cdots & b & a+\left(a^{n} b^{n}-a\right) \end{array}\right|\\ &=\left|\begin{array}{cccccc} a b & a & a & \cdots & a & 0 \\ b & a^{2} b^{2} & a & \cdots & a & 0 \\ b & b & a^{3} b^{3} & \cdots & a & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ b & b & b & \cdots & a^{n-1} b^{n-1} & 0 \\ b & b & b & \cdots & b & a^{n} b^{n}-a \end{array}\right|+\left|\begin{array}{cccccc} a b & a & a & \cdots & a & a \\ b & a^{2} b^{2} & a & \cdots & a & a \\ b & b & a^{3} b^{3} & \cdots & a & a \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ b & b & b & \cdots & a^{n-1} b^{n-1} & a \\ b & b & b & \cdots & b & a \end{array}\right|\\ &=\left(a^{n} b^{n}-a\right)\left|\begin{array}{ccclcc} a b & a & a & \cdots & a \\ b & a^{2} b^{2} & a & \cdots & a \\ b & b & a^{3} b^{3} & \cdots & a \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \cdots & a^{n-1} b^{n-1} \end{array}\right|+a\left|\begin{array}{cccccc} a b & a & a & \cdots & a & 1 \\ b & a^{2} b^{2} & a & \cdots & a & 1 \\ b & b & a^{3} b^{3} & \cdots & a & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ b & b & b & \cdots & a^{n-1} b^{n-1} & 1 \\ b & b & b & \cdots & b & 1 \end{array}\right|\\ &=\left(a^{n} b^{n}-a\right) D_{n-1}+a\left|\begin{array}{ccclcc} a b-b & a-b & a-b & \cdots & a-b & 1 \\ 0 & a^{2} b^{2}-b & a-b & \cdots & a-b & 1 \\ 0 & 0 & a^{3} b^{3}-b & \cdots & a-b & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a^{n-1} b^{n-1}-b & 1 \\ 0 & 0 & 0 & \cdots & 0 & 1 \end{array}\right|\\ &=\left(a^{n} b^{n}-a\right) D_{n-1}+a \prod\limits_{i=1}^{n-1}\left(a^{i} b^{i}-b\right). \end{aligned}$

由对称性可得 ${D_{n}=\left(a^{n} b^{n}-b\right) D_{n-1}+b \prod\limits_{i=1}^{n-1}\left(a^{i} b^{i}-a\right)}$ ,于是
${ \left\{\begin{array}{l} \left(a^{n} b^{n}-b\right) D_{n}=\left(a^{n} b^{n}-a\right)\left(a^{n} b^{n}-b\right) D_{n-1}+a \prod\limits_{i=1}^{n}\left(a^{i} b^{i}-b\right) \\ \left(a^{n} b^{n}-a\right) D_{n}=\left(a^{n} b^{n}-a\right)\left(a^{n} b^{n}-b\right) D_{n-1}+b \prod\limits_{i=1}^{n}\left(a^{i} b^{i}-a\right) \end{array}\right. }$
所以当 ${a \neq b}$ 时,上述两式相减可得
${ D_{n}=\frac{1}{b-a}\left[b \prod_{i=1}^{n}\left(a^{i} b^{i}-a\right)-a \prod_{i=1}^{n}\left(a^{i} b^{i}-b\right)\right] . }$
而明显 ${D_{n}}$ 是关于 ${a,b}$ 的连续函数,所以当 ${a=b}$ 时,结合数学分析中的洛必达法则,有
$\begin{aligned} D_{n} &=\lim _{b \rightarrow a} \frac{1}{b-a}\left[b \prod_{i=1}^{n}\left(a^{i} b^{i}-a\right)-a \prod_{i=1}^{n}\left(a^{i} b^{i}-b\right)\right] \\ &=\lim _{b \rightarrow a}\left[\prod_{i=1}^{n}\left(a^{i} b^{i}-a\right)+b \sum_{k=1}^{n}\left(i a^{i} b^{i-1} \prod_{i \neq k}\left(a^{i} b^{i}-a\right)\right)-a \sum_{k=1}^{n}\left(\left(i a^{i} b^{i-1}-1\right) \prod_{i \neq k}\left(a^{i} b^{i}-b\right)\right)\right] \\ &=\prod_{i=1}^{n}\left(a^{2 i}-a\right)+a \sum_{k=1}^{n}\left(i a^{2 i-1} \prod_{i \neq k}\left(a^{2 i}-a\right)\right)-a \sum_{k=1}^{n}\left(\left(i a^{2 i-1}-1\right) \prod_{i \neq k}\left(a^{2 i}-a\right)\right) \\ &=\prod_{i=1}^{n}\left(a^{2 i}-a\right)+a \sum_{k=1}^{n}\left(\prod_{i \neq k}\left(a^{2 i}-a\right)\right) \end{aligned}$

相邻行相消法

(合肥工业大学,2022)计算 ${n(n \geqslant 2)}$ 阶行列式
$D_{n}=\left|\begin{array}{cccccc} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{array}\right| .$

solution
从最后一行开始,上一行的 ${-1}$ 倍依次加到下一行,再将第一列的 ${-1}$ 倍加到之后每一列,最后将第 ${i(2 \leqslant i \leqslant n)}$ 列的 ${\dfrac{1}{n}}$ 倍均加到第一列,有
$\begin{aligned} D_{n}&=\left|\begin{array}{cccccc} 1 & 2 & 3 & \cdots & n-1 & n \\ 1 & 1 & 1 & \cdots & 1 & 1-n \\ 1 & 1 & 1 & \cdots & 1-n & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 1 & 1 & 1-n & \cdots & 1 & 1 \\ 1 & 1-n & 1 & \cdots & 1 & 1 \end{array}\right|\\&=\left|\begin{array}{cccccc} 1 & 1 & 2 & \cdots & n-2 & n-1 \\ 1 & 0 & 0 & \cdots & 0 & -n \\ 1 & 0 & 0 & \cdots & -n & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 1 & 0 & -n & \cdots & 0 & 0 \\ 1 & -n & 0 & \cdots & 0 & 0 \end{array}\right|\\ &=\left|\begin{array}{cccccc} 1+\sum\limits_{i=1}^{n-1} \dfrac{i}{n} & 1 & 2 & \cdots & n-2 & n-1 \\ 0 & 0 & 0 & \cdots & 0 & -n \\ 0 & 0 & 0 & \cdots & -n & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & -n & \cdots & -n & 0 \\ 0 & -n & 0 & \cdots & 0 & 0 \end{array}\right|\\ &=\dfrac{n+1}{2}(-1)^{\frac{(n-1)(n-2)}{2}}(-n)^{n-1}\\&=(-1)^{\frac{(n-1) n}{2}} \dfrac{n^{n-1}(n+1)}{2} \text {.} \end{aligned}$

(西南交通大学,2022;重庆大学,2022)计算行列式
$D_{n}=\left|\begin{array}{cccccc} x & 0 & 0 & \cdots & 0 & a_{0} \\ -1 & x & 0 & \cdots & 0 & a_{1} \\ 0 & -1 & x & \cdots & 0 & a_{2} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x & a_{n-2} \\ 0 & 0 & 0 & \cdots & -1 & x+a_{n-1} \end{array}\right|$

solution
从最后一行开始依次将每一行的 $x$ 倍加到上一行,然后按照第一行展开,就有
$\begin{aligned} D_{n}&=\left|\begin{array}{cccccc} 0 & 0 & 0 & \cdots & 0 & x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} \\ -1 & 0 & 0 & \cdots & 0 & x^{n-1}+a_{n-1} x^{n-2}+\cdots+a_{1} \\ 0 & -1 & 0 & \cdots & 0 & x^{n-2}+a_{n-1} x^{n-3}+\cdots+a_{2} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & x^{2}+a_{n-1} x+a_{n-2} \\ 0 & 0 & 0 & \cdots & -1 & x+a_{n-1} \end{array}\right|\\ &=\left(x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\right)(-1)^{n+1}\left|\begin{array}{cccc} -1 & & & \\ & -1 & & \\ & \ddots & \\ & & & -1 \end{array}\right|\\ &=\left(x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\right)(-1)^{n+1}(-1)^{n-1}\\&=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} . \end{aligned}$

计算行列式
$D_{n}=\left|\begin{array}{ccccc} 1 & 2 & 3 & \cdots & n \\ x & 1 & 2 & \cdots & n-1 \\ x & x & 1 & \cdots & n-2 \\ \vdots & \vdots & \vdots & & \vdots \\ x & x & x & \cdots & 1 \end{array}\right|$

solution
从第二行开始,每一行的一1倍依次加到上一行,然后按照最后一行拆为两个行列式,可得
$\begin{aligned} D_{n} &=\left|\begin{array}{cccccc} 1-x & 1 & 1 & \cdots & 1 & 1 \\ 0 & 1-x & 1 & \cdots & 1 & 1 \\ 0 & 0 & 1-x & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1-x & 1 \\ x & x & x & \cdots & x & 1 \end{array}\right|\\&=\left|\begin{array}{ccccccc} 1-x & 1 & 1 & \cdots & 1 & 1 \\ 0 & 1-x & 1 & \cdots & 1 & 1 \\ 0 & 0 & 1-x & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1-x & 1 \\ x & x & x & \cdots & x & x+(1-x) \end{array}\right| \\ &=\left|\begin{array}{cccccc} 1-x & 1 & 1 & \cdots & 1 & 1 \\ 0 & 1-x & 1 & \cdots & 1 & 1 \\ 0 & 0 & 1-x & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1-x & 1 \\ x & x & x & \cdots & x & x \end{array}\right|+\left|\begin{array}{cccccc} 1-x & 1 & 1 & \cdots & 1 & 1 \\ 0 & 1-x & 1 & \cdots & 1 & 1 \\ 0 & 0 & 1-x & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1-x & 1 \\ 0 & 0 & 0 & \cdots & 0 & 1-x \end{array}\right|\\ &=\left|\begin{array}{cccccc} -x & 0 & 0 & \cdots & 0 & 0 \\ -1 & -x & 0 & \cdots & 0 & 0 \\ -1 & -1 & -x & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ -1 & -1 & -1 & \cdots & -x & 0 \\ x & x & x & \cdots & x & x \end{array}\right|+(1-x)^{n}\\&=(-1)^{n-1} x^{n}+(1-x)^{n} . \end{aligned}$

高代 | 行列式 | 加边法与分解法 | 拆分法与递推法 | 相邻行相消法