行列式计算方法举例

1，化成上、下三角形行列式。

\begin{align*}|A|=\begin{vmatrix}1&1&1&\cdots&1\\ 1&0&1&\cdots&1\\1&1&0&\cdots&1\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&\cdots&0\end{vmatrix}=\begin{vmatrix}1&1&1&\cdots&1\\ 0&-1&0&\cdots&0\\0&0&-1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&-1\end{vmatrix}=(-1)^{n-1}\end{align*}

\begin{align*}|A|&=\begin{vmatrix}a_1&x&\cdots&x\\ x&a_2&\cdots&x\\ \vdots&\vdots&\ddots&\vdots\\ x&x&\cdots&a_n\end{vmatrix}=\begin{vmatrix}a_1&x&\cdots&x\\ x-a_1&a_2-x&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ x-a_1&0&\cdots&a_n-x\end{vmatrix}\end{align*}

\begin{align*}|A|&=\begin{vmatrix}a_1&x&\cdots&x\\ x-a_1&a_2-x&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ x-a_1&0&\cdots&a_n-x\end{vmatrix}=(a_1-x)(a_2-x)\cdots(a_n-x)\begin{vmatrix}\frac{a_1}{a_1-x}&\frac{x}{a_2-x}&\cdots&\frac{x}{a_n-x}\\ -1&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ -1&0&\cdots&1\end{vmatrix}\end{align*}

\begin{align*}|A|&=(a_1-x)(a_2-x)\cdots(a_n-x)\begin{vmatrix}\frac{a_1}{a_1-x}&\frac{x}{a_2-x}&\cdots&\frac{x}{a_n-x}\\ -1&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ -1&0&\cdots&1\end{vmatrix}\\ &=(a_1-x)(a_2-x)\cdots(a_n-x)\begin{vmatrix}\frac{a_1}{a_1-x}+\frac{x}{a_2-x}+\cdots+\frac{x}{a_n-x}&\frac{x}{a_2-x}&\cdots&\frac{x}{a_n-x}\\ 0&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&1\end{vmatrix}\\ &=(a_1-x)(a_2-x)\cdots(a_n-x)\cdot\left(\frac{a_1}{a_1-x}+\frac{x}{a_2-x}+\cdots+\frac{x}{a_n-x}\right)\end{align*}

2，行列式拆分。首先我们有下列的定理：

$|A|=\begin{vmatrix}a_{11}&\cdots&a_{1n}\\ \vdots&&\vdots\\ a_{i1}+b_1&\cdots&a_{in}+b_n\\ \vdots&&\vdots\\ a_{n1}&\cdots&a_{nn}\end{vmatrix}=\begin{vmatrix}a_{11}&\cdots&a_{1n}\\ \vdots&&\vdots\\ a_{i1}&\cdots&a_{in}\\ \vdots&&\vdots\\ a_{n1}&\cdots&a_{nn}\end{vmatrix}+\begin{vmatrix}a_{11}&\cdots&a_{1n}\\ \vdots&&\vdots\\ b_1&\cdots&b_n\\ \vdots&&\vdots\\ a_{n1}&\cdots&a_{nn}\end{vmatrix}$

$|A|=\begin{vmatrix}ax+by&ay+bz&az+bx\\ ay+bz&az+bx&ax+by\\ az+bx&ax+by&ay+bz\end{vmatrix}$ 的值。

\begin{align*}|A|&=\begin{vmatrix}ax+by&ay+bz&az+bx\\ ay+bz&az+bx&ax+by\\ az+bx&ax+by&ay+bz\end{vmatrix}\\&=\begin{vmatrix}ax&ay+bz&az+bx\\ ay&az+bx&ax+by\\ az&ax+by&ay+bz\end{vmatrix}+\begin{vmatrix}by&ay+bz&az+bx\\ bz&az+bx&ax+by\\ bx&ax+by&ay+bz\end{vmatrix}\\ &=a\begin{vmatrix}x&ay+bz&az+bx\\ y&az+bx&ax+by\\ z&ax+by&ay+bz\end{vmatrix}+b\begin{vmatrix}y&ay+bz&az+bx\\ z&az+bx&ax+by\\ x&ax+by&ay+bz\end{vmatrix}\end{align*}

\begin{align*}|A|&=a\begin{vmatrix}x&ay+bz&az+bx\\ y&az+bx&ax+by\\ z&ax+by&ay+bz\end{vmatrix}+b\begin{vmatrix}y&ay+bz&az+bx\\ z&az+bx&ax+by\\ x&ax+by&ay+bz\end{vmatrix}\\ &=a\begin{vmatrix}x&ay+bz&az\\ y&az+bx&ax\\ z&ax+by&ay\end{vmatrix}+b\begin{vmatrix}y&bz&az+bx\\ z&bx&ax+by\\ x&by&ay+bz\end{vmatrix}\\ &=a^2\begin{vmatrix}x&ay+bz&z\\ y&az+bx&x\\ z&ax+by&y\end{vmatrix}+b^2\begin{vmatrix}y&z&az+bx\\ z&x&ax+by\\ x&y&ay+bz\end{vmatrix} \\&=a^2\begin{vmatrix}x&ay&z\\ y&az&x\\ z&ax&y\end{vmatrix}+b^2\begin{vmatrix}y&z&bx\\ z&x&by\\ x&y&bz\end{vmatrix}\\&=a^3\begin{vmatrix}x&y&z\\y&z&x\\z&x&y\end{vmatrix}+b^3\begin{vmatrix}x&y&z\\y&z&x\\z&x&y\end{vmatrix}\\&=(a^3+b^3)D\end{align*}

3，行列式的递推法。有些行列式，低阶行列式与高阶行列式具有相同的形式，这时候，可以通过行列式的展开得到递推公式，利用递推公式，最后得到行列式的值。

\begin{align*}D_n&=(\alpha+\beta)\begin{vmatrix}\alpha+\beta& \alpha\beta&&&\\ 1&\alpha+\beta&\alpha\beta&&\\ 0&1&\alpha+\beta&&\\ \vdots&\vdots&\vdots&\ddots&\\ 0&0&0&\cdots&\alpha+\beta\end{vmatrix}-\alpha\beta\begin{vmatrix}1& \alpha\beta&&&\\ 0&\alpha+\beta&\alpha\beta&&\\ 0&1&\alpha+\beta&&\\ \vdots&\vdots&\vdots&\ddots&\\ 0&0&0&\cdots&\alpha+\beta\end{vmatrix}\\&= (\alpha+\beta)+\alpha\beta\begin{vmatrix}\alpha+\beta& \alpha\beta&&&\\ 1&\alpha+\beta&\alpha\beta&&\\ 0&1&\alpha+\beta&&\\ \vdots&\vdots&\vdots&\ddots&\\ 0&0&0&\cdots&\alpha+\beta\end{vmatrix}\\&=(\alpha+\beta)D_{n-1}-\alpha\beta D_{n-2}\end{align*}

\begin{align*}&D_n=\alpha D_{n-1}+\beta D_{n-1}-\alpha\beta D_{n-2}\\ \Rightarrow & D_n-\alpha D_{n-1}=\beta D_{n-1}-\alpha\beta D_{n-2}\\ \Rightarrow & D_n-\alpha D_{n-1}=\beta (D_{n-1}-\alpha D_{n-2})\end{align*}

\begin{align*}D_n-\alpha D_{n-1}&=\beta (D_{n-1}-\alpha D_{n-2})\\ &=\beta(\beta(D_{n-2}-\alpha D_{n-3}))\\ &=\beta^2(D_{n-2}-\alpha D_{n-3})=\cdots \\ &= \beta^{n-2}(D_2-\alpha D_1)\end{align*}

\begin{align*}D_{n}&=\alpha D_{n-1}+\beta^n=\alpha (\alpha D_{n-2}+\beta^{n-1})+\beta^n\\ &=\alpha^2 D_{n-2}+\alpha\beta^{n-1}+\beta^n\\ &=\alpha^2(\alpha D_{n-3}+\beta^{n-2})+\alpha\beta^{n-1}+\beta^n\\ &=\cdots=\alpha^{n-1}D_1+\alpha^{n-2}\beta^2+\cdots+\alpha\beta^{n-1}+\beta^n\\ &=\alpha^{n-1}(\alpha+\beta)+\alpha^{n-2}\beta^2+\cdots+\alpha\beta^{n-1}+\beta^n\\ &= \alpha^n+\alpha^{n-1}\beta+\alpha^{n-2}\beta^2+\cdots+\alpha\beta^{n-1}+\beta^n\\&=\begin{cases}(n+1)\alpha,\quad & \alpha=\beta,\\ \frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta},& \alpha\ne\beta\end{cases}\end{align*}