01-2行列式习题

Lingfeng 2025-07-15

01-2行列式习题

1. 常见性质

Theorem (鸡爪矩阵)

设 $A=\begin{pmatrix} a_{0} & 1 & 1 & \dots & 1 \\ 1 & a_{1} & 0& \dots & 0 \\ 1 & 0 & a_{2} & \dots & 0 \\ \vdots & \vdots &\vdots & & \vdots \\ 1 & 0 & 0 & \dots & a_{n} \end{pmatrix}$ 则 $\lvert A \rvert =a_{1}a_{2}\dots a_{n}\left( a_{0}-\sum_{i = 1}^{n} \frac{1}{a_{i}} \right)$

考虑初等变换 $C_{1} \leftarrow C_{1} - \frac{1}{a_{i}}C_{i+1},\quad i=1,\dots,n$ 此时 $\lvert A \rvert = \begin{pmatrix} a_{0}- \sum_{i = 1}^{n} \frac{1}{a_{i}} & 1 & 1 & \dots & 1 \\ 0 & a_{1} & 0& \dots & 0 \\ 0 & 0 & a_{2} & \dots & 0 \\ \vdots & \vdots &\vdots & & \vdots \\ 0 & 0 & 0 & \dots & a_{n} \end{pmatrix}$ 因此 $\lvert A \rvert = a_{1}a_{2}\dots a_{n}\left( a_{0}-\sum_{i = 1}^{n} \frac{1}{a_{i}} \right)$

Theorem (大拆分法)

设 $A=\left(\begin{matrix} a_{11}+x & a_{12}+x & \dots & a_{1n}+x \\ a_{21}+x & a_{22}+x & \dots & a_{2n}+x \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1}+x & a_{n2}+x & \dots & a_{nn}+x \end{matrix}\right)$ 则 $\lvert A \rvert =\left|\begin{matrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{matrix}\right|+ x\sum_{i = 1}^{n} \sum_{j = 1}^{n} A_{ij}$

对每一列进行拆分，因此 $\begin{align*} \lvert A \rvert & =\left|\begin{matrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{matrix}\right|+x\sum_{i = 1}^{n} \left|\begin{matrix} a_{11} & \dots & a_{1,i-1} & 1 & a_{1,i+1} & \dots & a_{1n} \\ a_{21} & \dots & a_{2,i-1} & 1 & a_{2,i+1} & \dots & a_{2n} \\ \vdots & & \vdots & \vdots & \vdots & & \vdots\\ a_{n1} & \dots & a_{n,i-1} & 1 & a_{n,i+1} & \dots & a_{nn} \\ \end{matrix}\right| \\ & = \left|\begin{matrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{matrix}\right| + x\sum_{i = 1}^{n} \sum_{j = 1}^{n} A_{ij} \end{align*}$

2. 课本习题

均选自第二章补充题。

Example (4.1)

计算 $\left|\begin{matrix} 1 & 2 & 3 & \dots & n \\ 2 & 3 & 4 & \dots & 1 \\ 3 & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ n & 1 & 2& \dots & n-1 \end{matrix}\right|$

考虑 $\begin{align*} d & = \left|\begin{matrix} \frac{n(n+1)}{2} & 2 & 3 & \dots & n \\ \frac{n(n+1)}{2} & 3 & 4 & \dots & 1 \\ \frac{n(n+1)}{2} & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ \frac{n(n+1)}{2}n & 1 & 2& \dots & n-1 \end{matrix}\right| = \frac{n(n+1)}{2} \left|\begin{matrix} 1 & 2 & 3 & \dots & n \\ 1 & 3 & 4 & \dots & 1 \\ 1 & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ 1 & 1 & 2& \dots & n-1 \end{matrix}\right| \\ & = \frac{n(n+1)}{2}\left|\begin{matrix} 1 & 2 & 3 & \dots & n \\ 0 & 1 & 1 & \dots & 1-n \\ 0 & 1 & 1 & \dots & 1 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 1-n & 1& \dots & 1 \end{matrix}\right| = \frac{n(n+1)}{2}\left|\begin{matrix} 1 & 1 & \dots & 1 & 1-n \\ 1 & 1 & \dots & 1-n & 1 \\ \vdots & \vdots & & \vdots & \vdots \\ 1-n & 1& \dots & 1 & 1\end{matrix}\right| \\ & = \frac{n(n-1)}{2}d_{1} \end{align*}$ 下面使用大拆分法，有
$\begin{align*} d_{1}& = (-1)^{\frac{n(n-1)}{2}}((-n)^{n-1} + (n-1)(-n)^{n-2}) \\ & = (-1)^{\frac{n(n-1)}{2}}(-1)^{n-2}n^{n-2}(-n+n-1) \\ & = (-1)^{\frac{(n+2)(n-1)}{2}} n^{n-2} \end{align*}$ 因此 $d=(-1)^{\frac{(n+2)(n-1)}{2}}\frac{n^{n-1}(n-1)}{2}$

Example (4.2)

计算 $\left|\begin{matrix} \lambda & a & a & a& \dots & a \\ b & \alpha & \beta & \beta & \dots & \beta \\ b & \beta & \alpha & \beta & \dots & \beta \\ b & \beta & \beta & \alpha & \dots & \beta \\ \vdots & \vdots & \vdots & \vdots & & \vdots\\ b & \beta & \beta & \beta & \dots & \alpha\end{matrix}\right|$

考虑小拆分法 $\begin{align*} d & = \left|\begin{matrix} \lambda-b & a & a & a& \dots & a \\ 0 & \alpha & \beta & \beta & \dots & \beta \\ 0 & \beta & \alpha & \beta & \dots & \beta \\ 0 & \beta & \beta & \alpha & \dots & \beta \\ \vdots & \vdots & \vdots & \vdots & & \vdots\\ 0 & \beta & \beta & \beta & \dots & \alpha\end{matrix}\right| + b\left|\begin{matrix} 1 & a & a & a& \dots & a \\ 1 & \alpha & \beta & \beta & \dots & \beta \\ 1 & \beta & \alpha & \beta & \dots & \beta \\ 1 & \beta & \beta & \alpha & \dots & \beta \\ \vdots & \vdots & \vdots & \vdots & & \vdots\\ 1 & \beta & \beta & \beta & \dots & \alpha\end{matrix}\right| \\ & =(\lambda-b) \left|\begin{matrix} \alpha & \beta & \beta & \dots & \beta \\ \beta & \alpha & \beta & \dots & \beta \\ \beta & \beta & \alpha & \dots & \beta \\ \vdots & \vdots & \vdots & & \vdots\\ \beta & \beta & \beta & \dots & \alpha\end{matrix}\right| + b \left|\begin{matrix} 1 & a & a & a& \dots & a \\ 0 & \alpha -a & \beta-a & \beta-a & \dots & \beta-a \\ 0 & \beta-\alpha & \alpha-\beta & 0 & \dots & 0 \\ 0 & 0 & \beta-\alpha & \alpha-\beta & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots\\ 0 & 0 & 0 & 0 & \dots & \alpha-\beta\end{matrix}\right| \\ & = (\lambda-b) d_{1}+bd_{2} \end{align*}$ $d_{1}$ 使用大拆分法 $d_{1} = (\alpha-\beta)^{n-1}+(n-1)\beta(\alpha-\beta)^{n-2}=(\alpha-\beta)^{n-2}(\alpha+(n-2)\beta)$ $d_{2}$ 全加到第一列 $\begin{align*} d_{2} & = \left|\begin{matrix} \alpha -a & \beta-a & \beta-a & \dots & \beta-a \\ \beta-\alpha & \alpha-\beta & 0 & \dots & 0 \\ 0 & \beta-\alpha & \alpha-\beta & \dots & 0 \\ \vdots & \vdots & \vdots & & \vdots\\ 0 & 0 & 0 & \dots & \alpha-\beta\end{matrix}\right| \\ & = \left|\begin{matrix} (\alpha -a) + (n-2)(\beta-a) & \beta-a & \beta-a & \dots & \beta-a \\ 0 & \alpha-\beta & 0 & \dots & 0 \\ 0 & \beta-\alpha & \alpha-\beta & \dots & 0 \\ \vdots & \vdots & \vdots & & \vdots\\ 0 & 0 & 0 & \dots & \alpha-\beta\end{matrix}\right| \\ & = ((\alpha-a)+(n-2)(\beta-a))(\alpha-\beta)^{n-2} \\ & = (\alpha-\beta)^{n-2}(\alpha+(n-2)\beta-(n-1)a) \end{align*}$ 因此 $\begin{align*} d & =(\lambda-b) d_{1}+bd_{2} \\ & = (\alpha-\beta)^{n-2}((\lambda\alpha+(n-2)\lambda\beta)-(n-1)ab) \end{align*}$

Example (4.3)

计算 $\left | \begin{matrix} x & a & a & \dots & a & a \\ -a & x & a & \dots & a & a \\ -a & -a & x & \dots & a & a \\ \vdots & \vdots &\vdots & & \vdots & \vdots \\ -a & -a & -a & \dots & -a & x\end{matrix} \right |$

同样小拆分法 $\begin{align*} d_{n} & = \left | \begin{matrix} x +a & a & a & \dots & a & a \\ 0 & x & a & \dots & a & a \\ 0 & -a & x & \dots & a & a \\ \vdots & \vdots &\vdots & & \vdots & \vdots \\ 0 & -a & -a & \dots & -a & x\end{matrix} \right| -a \left | \begin{matrix} 1 & a & a & \dots & a & a \\ 1 & x & a & \dots & a & a \\ 1 & -a & x & \dots & a & a \\ \vdots & \vdots &\vdots & & \vdots & \vdots \\ 1 & -a & -a & \dots & -a & x\end{matrix} \right| \\ & = (x+a) d_{n-1} - a \left | \begin{matrix} x-a & 0 & \dots & 0 & 0 \\ -a-x & x-a & \dots & 0 & 0 \\ \vdots &\vdots & & \vdots & \vdots \\ 0 & 0 & \dots & -a-x & x-a\end{matrix} \right| \\ & = (x+a)d_{n-1}-a(x-a)^{n-1} \end{align*}$ 下面考虑数列通项，注意到 $\frac{d_{n}}{(x-a)^{n}}=\frac{x+a}{x-a} \cdot \frac{d_{n-1}}{(x-a)^{n-1}}- \frac{a}{x-a}$ 令 $c_{n}=\frac{d_{n}}{(x-a)^{n}}$ ，此时 $c_{n}-\frac{1}{2}= \frac{x+a}{x-a} \left( c_{n-1} - \frac{1}{2} \right)$ 而 $c_{1}-\frac{1}{2}=\frac{x}{x-a}-\frac{1}{2}=\frac{x+a}{2(x-a)}$ 因此 $\frac{d_{n}}{(x-a)^{n}}=\left( \frac{x+a}{x-a} \right)^{n-1}\frac{x+a}{2(x-a)} + \frac{1}{2}$ 即
$d_{n}=\frac{1}{2} ((x+a)^{n}+ (x-a)^{n})$

Example (4.4)

计算 $\left | \begin{matrix} x & y & y & \dots & y & y \\ z & x & y & \dots & y & y \\ z & z & x & \dots & y & y\\ \vdots & \vdots & \vdots & & \vdots & \vdots\\ z & z & z & \dots & x & y\\ z & z & z & \dots & z & x\end{matrix} \right |$

跟4.3完全一致，这里使用另一种方法

第一列小拆分法 $\begin{align*} d_{n} & = (x -z)\left | \begin{matrix} x & y & \dots & y & y \\ z & x & \dots & y & y\\ \vdots & \vdots & & \vdots & \vdots\\ z & z & \dots & x & y\\ z & z & \dots & z & x\end{matrix} \right | + z \left | \begin{matrix} 1 & y & y & \dots & y & y \\ 1 & x & y & \dots & y & y \\ 1 & z & x & \dots & y & y\\ \vdots & \vdots & \vdots & & \vdots & \vdots\\ 1 & z & z & \dots & x & y\\ 1 & z & z & \dots & z & x\end{matrix} \right | \\ & = (x-z)d_{n-1} + z \left | \begin{matrix} x-y & 0 & \dots & 0 & 0 \\ z-x & x-y & \dots & 0 & 0\\ \vdots & \vdots & & \vdots & \vdots\\ 0 & 0 & \dots & x-y & 0\\ 0 & 0 & \dots & z-x & x-y\end{matrix} \right | \\ & = (x-z)d_{n-1} + z(x-y)^{n-1} \end{align*}$ 我们再对最后一列进行小拆分 $\begin{align*} d_{n} & = (x-y)\left | \begin{matrix} x & y & y & \dots & y \\ z & x &y & \dots & y \\ z & z & x & \dots & y \\ \vdots & \vdots & \vdots & & \vdots\\ z & z & z & \dots & x \end{matrix} \right | + y\left | \begin{matrix} x & y & y & \dots & y & 1 \\ z & x & y & \dots & y & 1 \\ z & z & x & \dots & y & 1\\ \vdots & \vdots & \vdots & & \vdots & \vdots\\ z & z & z & \dots & x & 1\\ z & z & z & \dots & z & 1\end{matrix} \right | \\ & = (x-y)d_{n-1} + y \left | \begin{matrix} x-z & y-x & 0 & \dots & 0 \\ 0 & x-z &y-x & \dots & 0 \\ 0 & 0 & x-z & \dots & 0 \\ \vdots & \vdots & \vdots & & \vdots\\ 0 & 0 & 0 & \dots & x-z \end{matrix} \right | \\ & =(x-y)d_{n-1}+y(x-z)^{n-1} \end{align*}$ 联立即可解出 $d_{n}=\frac{1}{z-y}(z(x-y)^{n}-y(x-z)^{n})$

Example (4.5)

计算 $\left | \begin{matrix} 1 & 1 & \dots & 1 \\ x_{1} & x_{2} & \dots & x_{n} \\ x_{1}^{2} & x_{2}^{2} & \dots & x_{n}^{2} \\ \vdots & \vdots & & \vdots \\ x_{1}^{n-2} & x_{2}^{n-2} & \dots & x_{n}^{n-2} \\ x_{1}^{n} & x_{2}^{n} & \dots & x_{n}^{n}\end{matrix} \right |$

联想范德蒙德行列式，构造 $P(t)=\left | \begin{matrix} 1 & 1 & \dots & 1 & 1 \\ x_{1} & x_{2} & \dots & x_{n} & t \\ x_{1}^{2} & x_{2}^{2} & \dots & x_{n}^{2} & t^{2}\\ \vdots & \vdots & & \vdots & \vdots\\ x_{1}^{n-2} & x_{2}^{n-2} & \dots & x_{n}^{n-2} & t^{n-2}\\ x_{1}^{n-1} & x_{2}^{n-1} & \dots & x_{n}^{n-1} & t^{n-1} \\ x_{1}^{n} & x_{2}^{n} & \dots & x_{n}^{n} & t^{n}\end{matrix} \right |$ 此时 $P(t)=\prod_{1\leq j<i\leq n}(x_{i}-x_{j}) \prod_{i=1}^{n}(t-x_{i})$ 注意到题意所求即 $t^{n-1}$ 展开的余子式，因此考虑 $t^{n-1}$ 系数 $(-1)^{2n+1}d=-\prod_{1\leq j<i\leq n}(x_{i}-x_{j}) \sum_{i = 1}^{n} x_{i}$ 即 $d=\prod_{1\leq j<i\leq n}(x_{i}-x_{j}) \sum_{i = 1}^{n} x_{i}$