1.典错题

标签：right ddots ij 错题 ab left vert

逆矩阵

因式分解求逆矩阵

行列式

范德蒙德行列式

每行/列元素之和相等

递推/归纳

1.

2.

\(\begin{vmatrix} a+b & ab & 0 & \cdots & 0\\ 1 & a+b & ab & \cdots & 0\\ \\ & \ddots & \ddots & \ddots\\ \\ & &\ddots & a+b & ab \\ & & & 1 & a+b \end{vmatrix}=?\)
解:

\[\begin{aligned} 第一行展开\\D_{n}=(a+b)D_{n_-1}-abD_{n-2}\\D_{n}-bD_{n-1}=a(D_{n-1}-bD_{n-2}) \\D_1=a+b,D_2=(a+b)^{2}-ab \\ D_n-bD_{n-1}=a^{n-2}(a^{2})=a^{n} \\ 同理,D_n-aD_{n-1}=b^{n} \\ \therefore D_n=\frac{a^{n+1}-b^{n+1}}{a-b}=a^{n}+a^{n-1}b+ \cdots + b^{n} \end{aligned} \]

杂项

1.

\[A_{3\times 3} ,已知a_{ij}+A_{ij}=0,则\left\vert A \right\vert=? \]

解:

\[a_{ij}+A_{ij}=0 \rightarrow A^{T}=-A^{*}\rightarrow\left\vert A \right\vert =\left\vert A^{T} \right\vert =-\left\vert A^{*} \right\vert =-\left\vert A \right\vert ^{2} \therefore \left\vert A \right\vert =0/ -1 \]

\[\because \left\vert A \right\vert= a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}=-(a_{11}^{2}+a_{12}^{2}+a_{13}^{2})\neq 0 \]

\[\therefore \left\vert A \right\vert=-1 \]

标签：right,ddots,ij,错题,ab,left,vert
From： https://www.cnblogs.com/zjxxinseu/p/16831925.html