要想及时的了解数学信息,请中学师生关注公众号“中学究尽数学”;大学生关注公众号“究尽数学”

行列式的计算方法专题小结

数学心得 huijiaorz 114℃ 0评论

熟练掌握行列式的性质,再结合一些技巧来计算行列式。该文简单介绍几种行列式的计算方法:通过按行(列)展开,或者通过添加行列使得行列式转化为递归式,此时可通过简单递归或者联立等式来求得原行列式;通过观察行列式,如果行和或列和相等的话,可以通过提出行和,再消去一行一列;将一个行列式分解成两个行列式的乘积,或者两个行列式的和,再结合行列式的性质及其他计算方法分别求出行列式;最后的参数法也是非常的巧妙,会将行列式转化为函数问题。行列式的计算技巧性很强,平时多积累,是非常好的思维训练素材。

按某一行(列)展开

计算行列式 (Hessenberg型行列式)$$\mathbf{D_n}=\left|\begin{array}{cccccc}x&-1&0&\ldots&0&0\\0&x&-1&\ldots&0&0\\0&0&x&\ldots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\ldots&x&-1\\a_n&a_{n-1}&a_{n-2}&\ldots&a_2&x+a_1\\\end{array}\right|$$

解析:按第一列展开$$\begin{aligned}\mathbf{D_n}&=x\cdot\left|\begin{array}{cccccc}x&-1&\ldots&0&0\\0&x&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&x&-1\\a_{n-1}&a_{n-2}&\ldots&a_2&x+a_1\\\end{array}\right|+a_n\cdot(-1)^{n+1}(-1)^{n-1}\\&=x\cdot{\mathbf{D_{n-1}}}+a_n\\&=x[x\cdot{\mathbf{D_{n-2}}}+a_{n-1}]+a_n\\&=x^2\mathbf{D_{n-2}}+a_{n-1}x+a_n\\&=\cdots\\&=x^{n-2}\mathbf{D_2}+a_3x^{n-3}+a_4x^{n-4}+\cdots+a_{n-1}x+a_n\\\\\mathbf{D_2}&=\left|\begin{array}{cc}x&-1\\a_2&x+a_1\\\end{array}\right|\\&=x^2+a_1x+a_2\end{aligned}$$

所以$$\mathbf{D_n}=x^n+a_1x^{n-1}+a_2x^{n-2}+a_3x^{n-3}+a_4x^{n-4}+\cdots+a_{n-1}x+a_n$$

添加一行(列)

计算行列式:$$\left|\begin{array}{cccc}1+x_1^2&x_1x_2&\ldots&x_1x_n\\x_2x_1&1+x_2^2&\ldots&x_2x_n\\\vdots&\vdots&\ddots&\vdots\\x_nx_1&x_nx_2&\ldots&1+x_n^2\\\end{array}\right|$$

解:添加一行一列$$\begin{aligned}D_n&=\left|\begin{array}{ccccc}1&x_1&x_2&\ldots&x_n\\0&1+x_1^2&x_1x_2&\ldots&x_1x_n\\0&x_2x_1&1+x_2^2&\ldots&x_2x_n\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&x_nx_1&x_nx_2&\ldots&1+x_n^2\\\end{array}\right|\\&=\left|\begin{array}{ccccc}1&x_1&x_2&\ldots&x_n\\-x_1&1&0&\ldots&0\\-x_2&0&1&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\-x_n&0&0&\ldots&1\\\end{array}\right|\\&=\left|\begin{array}{ccccc}1+x_1^2+x_2^2+\cdots+x_n^2&x_1&x_2&\ldots&x_n\\0&1&0&\ldots&0\\0&0&1&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&1\\\end{array}\right|\\&=1+\sum_{i=1}^nx_i^2\end{aligned}$$

该行列式也可以通过拆分为两个行列式之和,然后分别求两个行列式。可参考“行列式的拆分”

逐行(列)求和

计算行列式:$$\left|\begin{array}{cccc}1+x&2&\ldots&n\\1&2+x&\ldots&n\\\vdots&\vdots&\ddots&\vdots\\1&2&\ldots&n+x\\\end{array}\right|$$

解:$$\begin{aligned}\left|\begin{array}{cccc}1+x&2&\ldots&n\\1&2+x&\ldots&n\\\vdots&\vdots&\ddots&\vdots\\1&2&\ldots&n+x\\\end{array}\right|&=\left|\begin{array}{cccc}x+\sum_{i=1}^ni&2&\ldots&n\\x+\sum_{i=1}^ni&2+x&\ldots&n\\\vdots&\vdots&\ddots&\vdots\\x+\sum_{i=1}^ni&2&\ldots&n+x\\\end{array}\right|\\&=\left|\begin{array}{cccc}1&2&\ldots&n\\1&2+x&\ldots&n\\\vdots&\vdots&\ddots&\vdots\\1&2&\ldots&n+x\\\end{array}\right|\cdot[x+\frac{n(n+1)}{2}]\\&=\left|\begin{array}{cccc}1&0&\ldots&0\\1&x&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\1&0&\ldots&x\\\end{array}\right|\cdot[x+\frac{n(n+1)}{2}]\\&=[x+\frac{n(n+1)}{2}]\cdot{x^{n-1}}\\&=x^n+\frac{n^2+n}{2}x^{n-1}\end{aligned}$$

行列式的“因式分解”

计算行列式:$$\left|\begin{array}{cccc}a_1-b_1&a_1-b_2&\ldots&a_1-b_n\\a_2-b_1&a_2-b_2&\ldots&a_2-b_n\\\vdots&\vdots&\ddots&\vdots\\a_n-b_1&a_n-b_2&\ldots&a_n-b_n\\\end{array}\right|$$

解:$$\begin{aligned}D&=\left|\begin{array}{cccc}a_1-b_1&a_1-b_2&\ldots&a_1-b_n\\a_2-b_1&a_2-b_2&\ldots&a_2-b_n\\\vdots&\vdots&\ddots&\vdots\\a_n-b_1&a_n-b_2&\ldots&a_n-b_n\\\end{array}\right|\\&=\left|\begin{array}{cccc}a_1&-1&\ldots&0\\a_2&-1&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\a_n&-1&\ldots&0\\\end{array}\right|\cdot\left|\begin{array}{cccc}1&1&\ldots&1\\b_1&b_2&\ldots&b_n\\\vdots&\vdots&\ddots&\vdots\\0&0&\ldots&0\\\end{array}\right|\\&=\begin{cases}(a_2-a_1)(b_2-b_1),\quad n=2\\0,\quad n>2\end{cases}\end{aligned}$$

设$S_k=x_1^k+x_2^k+\cdots+x_n^k$,其中$k$为任意非负整数,证明:$$D=\left|\begin{array}{cccc}S_0&S_1&\ldots&S_{n-1}\\S_1&S_2&\ldots&S_n\\\vdots&\vdots&\ddots&\vdots\\S_{n-1}&S_n&\ldots&S_{2n-2}\\\end{array}\right|=\prod_{1\leq i< j\leq n}(x_i-x_j)^2$$

证明:$$\begin{aligned}D&=\left|\begin{array}{cccc}1&1&\ldots&1\\x_1&x_2&\ldots&x_n\\\vdots&\vdots&\ddots&\vdots\\x_1^{n-1}&x_2^{n-1}&\ldots&x_n^{n-1}&\end{array}\right|\cdot\left|\begin{array}{cccc}1&x_1&\ldots&x_1^{n-1}\\1&x_2&\ldots&x_2^{n-1}\\\vdots&\vdots&\ddots&\vdots\\1&x_n&\ldots&x_n^{n-1}&\end{array}\right|\\&=\prod_{1\leq i< j\leq n}(x_j-x_i)\cdot\prod_{1\leq i< j\leq n}(x_j-x_i)\end{aligned}$$

行列式的拆分

计算行列式:$$\left|\begin{array}{ccccc}x&b&b&\ldots&b\\c&x&b&\ldots&b\\c&c&x&\ldots&b\\\vdots&\vdots&\vdots&\ddots&\vdots\\c&c&c&\ldots&x\\\end{array}\right|$$

解:$$\begin{aligned}D_n&=\left|\begin{array}{ccccc}x&b&b&\ldots&b\\c&x&b&\ldots&b\\c&c&x&\ldots&b\\\vdots&\vdots&\vdots&\ddots&\vdots\\c&c&c&\ldots&x\\\end{array}\right|\\&=\left|\begin{array}{ccccc}c+(x-c)&b&b&\ldots&b\\c&x&b&\ldots&b\\c&c&x&\ldots&b\\\vdots&\vdots&\vdots&\ddots&\vdots\\c&c&c&\ldots&x\\\end{array}\right|\\&=c\cdot\left|\begin{array}{ccccc}1&b&b&\ldots&b\\1&x&b&\ldots&b\\1&c&x&\ldots&b\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&c&c&\ldots&x\\\end{array}\right|+\left|\begin{array}{ccccc}x-c&b&b&\ldots&b\\0&x&b&\ldots&b\\0&c&x&\ldots&b\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&c&c&\ldots&x\\\end{array}\right|\\&=c\cdot\left|\begin{array}{ccccc}1&0&0&\ldots&0\\0&x-b&0&\ldots&0\\0&c-b&x-b&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&c-b&c-b&\ldots&x-b\\\end{array}\right|+(x-c)\cdot D_{n-1}\\&=c(x-b)^{n-1}+(x-c)D_{n-1}\end{aligned}$$

同理可得$$D_n=b(x-c)^{n-1}+(x-b)D_{n-1}$$

联立两个等式,消去$D_{n-1}$得$$(b-c)\cdot D_n=b(x-c)^n-c(x-b)^n$$

所以当$b\neq c$时$$D_n=\frac{b(x-c)^n-c(x-b)^n}{b-c}$$

而$b=c$时$$D_n=[x+(n-1)b](x-b)^{n-1}$$

计算行列式:$$\left|\begin{array}{cccc}1+x_1^2&x_1x_2&\ldots&x_1x_n\\x_2x_1&1+x_2^2&\ldots&x_2x_n\\\vdots&\vdots&\ddots&\vdots\\x_nx_1&x_nx_2&\ldots&1+x_n^2\\\end{array}\right|$$

解:$$\begin{aligned}D_n&=\left|\begin{array}{cccc}1+x_1^2&x_1x_2&\ldots&0\\x_2x_1&1+x_2^2&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\ldots&1\\\end{array}\right|+\left|\begin{array}{cccc}1+x_1^2&x_1x_2&\ldots&x_1x_n\\x_2x_1&1+x_2^2&\ldots&x_2x_n\\\vdots&\vdots&\ddots&\vdots\\x_nx_1&x_nx_2&\ldots&x_n^2\\\end{array}\right|\\&=D_{n-1}+x_n^2\left|\begin{array}{cccc}1+x_1^2&x_1x_2&\ldots&x_1\\x_2x_1&1+x_2^2&\ldots&x_2\\\vdots&\vdots&\ddots&\vdots\\x_1&x_2&\ldots&1\\\end{array}\right|\\&=D_{n-1}+x_n^2\left|\begin{array}{cccc}1&0&\ldots&0\\0&1&\ldots&0\\\vdots&\vdots&\ddots&\vdots\\x_1&x_2&\ldots&1\\\end{array}\right|\\&=D_{n-1}+x_n^2\\&=\cdots\\&=1+x_1^2+x_2^2+\cdots+x_n^2\end{aligned}$$

参数法

求$$D_n=\left|\begin{array}{ccccc}x_1&a&\ldots&a&a\\b&x_2&\ldots&a&a\\\vdots&\vdots&\ddots&\vdots&\vdots\\b&b&\ldots&x_{n-1}&a\\b&b&\ldots&b&x_n\\\end{array}\right|$$

解:记$$D_n(t)=\left|\begin{array}{ccccc}x_1+t&a+t&\ldots&a+t&a+t\\b+t&x_2+t&\ldots&a+t&a+t\\\vdots&\vdots&\ddots&\vdots&\vdots\\b+t&b+t&\ldots&x_{n-1}+t&a+t\\b+t&b+t&\ldots&b+t&x_n+t\\\end{array}\right|=D_n+t\cdot r$$

其中$t$与$r$无关。分别取$t=-a$和$t=-b$,可得:$$D_n(-a)=D_n-a\cdot r=\prod_{i=1}^n(x_i-a)\\D_n(-b)=D_n-b\cdot r=\prod_{i=1}^n(x_i-b)$$

两式联立消去$r$可得$$D_n=\frac{a\prod_{i=1}^n(x_i-b)-b\prod_{i=1}^n(x_i-a)}{a-b}$$

转载请注明:究尽数学 » 行列式的计算方法专题小结

喜欢 (1)or分享 (0)
发表我的评论
取消评论
表情

Hi,您需要填写昵称和邮箱!

  • 昵称 (必填)
  • 邮箱 (必填)
  • 网址