范德蒙行列式“变体”行列式的计算

一、前言

$$D _{ n } = \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ x _{ 1 } & x _{ 2 } & x _{ 3 } & \cdots & x _{ n } \\ x _{ 1 } ^ { 2 } & x _{ 2 } ^ { 2 } & x _{ 3 } ^ { 2 } & \cdots & x _{ n } ^ { 2 } \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ x _{ 1 } ^ { n – 1 } & x _{ 2 } ^ { n – 1 } & x _{ 3 } ^ { n – 1 } & \cdots & x _{ n } ^ { n – 1 } \end{vmatrix}$$

$$D _{ n } = \prod _{ 1 \leqslant j < i \leqslant n } \left( x _{ i } – x _{ j } \right)$$

二、正文

§1.2 解析一

\begin{aligned} & |K_{1}| \\ \\ = & \begin{vmatrix} 1 & a & b c \\ 1 & b & a c \\ 1 & c & a b \end{vmatrix} \\ \\ \xlongequal{第 \ 1 \ 行乘以 \ a} \frac{1}{a} & \begin{vmatrix} a & a^{2} & a b c \\ 1 & b & a c \\ 1 & c & a b \end{vmatrix} \\ \\ \xlongequal{第 \ 2 \ 行乘以 \ b} \frac{1}{ab} & \begin{vmatrix} a & a^{2} & a b c \\ b & b^{2} & a b c \\ 1 & c & a b \end{vmatrix} \\ \\ \xlongequal{第 \ 2 \ 行乘以 \ c} \frac{1}{abc} & \begin{vmatrix} a & a^{2} & \textcolor{pink}{a b c} \\ b & b^{2} & \textcolor{pink}{a b c} \\ c & c^{2} & \textcolor{pink}{a b c} \end{vmatrix} \\ \\ \xlongequal{提取第 \ 3 \ 列的 \ \textcolor{pink}{abc}} \frac{1}{abc} \cdot \textcolor{pink}{abc} & \begin{vmatrix} a & a^{2} & \textcolor{pink}{1} \\ b & b^{2} & \textcolor{pink}{1} \\ c & c^{2} & \textcolor{pink}{1} \end{vmatrix} \\ \\ \xlongequal{第 \ 1 \ 列和第 \ 3 \ 列交换} (-1) \cdot & \begin{vmatrix} 1 & a^{2} & a \\ 1 & b^{2} & b \\ 1 & c^{2} & c \end{vmatrix} \\ \\ \xlongequal{第 \ 2 \ 列和第 \ 3 \ 列交换} (-1) \cdot (-1) \cdot & \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix} \\ \\ = & \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix} \\ \\ \xlongequal{转置不会改变行列式的值} & \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{vmatrix} \\ \\ \xlongequal{范德蒙行列式计算公式} & \textcolor{springgreen}{\boldsymbol{(b-a)(c-a)(c-b)}} \end{aligned}

§2.2 解析二

\begin{aligned} & |K_{2}| \\ \\ = & \begin{vmatrix} 1 & a & \textcolor{springgreen}{b c} + \textcolor{magenta}{a ^ { 2 }} \\ 1 & b & \textcolor{springgreen}{a c} + \textcolor{magenta}{b ^ { 2 }} \\ 1 & c & \textcolor{springgreen}{a b} + \textcolor{magenta}{c ^ { 2 }} \end{vmatrix} \\ \\ = & \begin{vmatrix} 1 & a & \textcolor{springgreen}{b c} \\ 1 & b & \textcolor{springgreen}{a c} \\ 1 & c & \textcolor{springgreen}{a b} \end{vmatrix} + \begin{vmatrix} 1 & a & \textcolor{magenta}{a ^ { 2 }} \\ 1 & b & \textcolor{magenta}{b ^ { 2 }} \\ 1 & c & \textcolor{magenta}{c ^ { 2 }} \end{vmatrix} \end{aligned}

$$\begin{vmatrix} 1 & a & \textcolor{springgreen}{b c} \\ 1 & b & \textcolor{springgreen}{a c} \\ 1 & c & \textcolor{springgreen}{a b} \end{vmatrix} = \textcolor{springgreen}{(b-a)(c-a)(c-b)}$$

$$\begin{vmatrix} 1 & a & \textcolor{magenta}{a ^ { 2 }} \\ 1 & b & \textcolor{magenta}{b ^ { 2 }} \\ 1 & c & \textcolor{magenta}{c ^ { 2 }} \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ \textcolor{magenta}{a ^{2}} & \textcolor{magenta}{b ^{2}} & \textcolor{magenta}{c ^{2}} \end{vmatrix} = \textcolor{springgreen}{(b-a)(c-a)(c-b)}$$

$$|K_{2}| = \textcolor{springgreen}{2 \cdot \boldsymbol{(b-a)(c-a)(c-b)}}$$

§3.2 解析三

\begin{aligned} & |K_{3}| \\ \\ = & \begin{vmatrix} b + c + d & a + c + d & a + b + d & a + b + c \\ \textcolor{springgreen}{a} & \textcolor{pink}{b} & \textcolor{orangered}{c} & \textcolor{yellow}{d} \\ a ^ { 2 } & b ^ { 2 } & c ^ { 2 } & d ^ { 2 } \\ a ^ { 3 } & b ^ { 3 } & c ^ { 3 } & d ^ { 3 } \end{vmatrix} \\ \\ = & \begin{vmatrix} \textcolor{springgreen}{a} + b + c + d & a + \textcolor{pink}{b} + c + d & a + b + \textcolor{orangered}{c} + d & a + b + c + \textcolor{yellow}{d} \\ \textcolor{springgreen}{a} & \textcolor{pink}{b} & \textcolor{orangered}{c} & \textcolor{yellow}{d} \\ a ^ { 2 } & b ^ { 2 } & c ^ { 2 } & d ^ { 2 } \\ a ^ { 3 } & b ^ { 3 } & c ^ { 3 } & d ^ { 3 } \end{vmatrix} \\ \\ = & (a+b+c+d) \cdot \begin{vmatrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a ^ { 2 } & b ^ { 2 } & c ^ { 2 } & d ^ { 2 } \\ a ^ { 3 } & b ^ { 3 } & c ^ { 3 } & d ^ { 3 } \end{vmatrix} \\ \\ = & \textcolor{springgreen}{\boldsymbol{ (a+b+c+d) \cdot (b-a) (c-a) (d-a) (c-b) (d – b) (d -c) }} \end{aligned}

§4.2 解析四

\begin{aligned} & \begin{cases} (-2) ^{2} = 4 \\ (-3) ^{2} = 9 \\ \end{cases} \\ \\ & \begin{cases} (-2) ^{3} = -8 \\ (-3) ^{3} = -27 \end{cases} \end{aligned}

\begin{aligned} & |K_{4}| \\ \\ = & \begin{vmatrix} 1 & 1 & 1 & 1 \\ – 2 & – 3 & – 4 & – 5 \\ 4 & 9 & 16 & 25 \\ – 8 & – 27 & – 64 & – 125 \end{vmatrix} \\ \\ = & \begin{vmatrix} 1 & 1 & 1 & 1 \\ – 2 & – 3 & – 4 & – 5 \\ (-2) ^{2} & (-3) ^{2} & (-4) ^{2} & (-5) ^{2} \\ (-2) ^{3} & (-3) ^{3} & (-4) ^{3} & (-5) ^{3} \end{vmatrix} \\ \\ = & (-3+2) (-4 + 2) (-5 + 2) (-4+3) (-5+3) (-5+4) \\ \\ = & (-1) (-2) (-3) (-1) (-2) (-1) \\ \\ = & (-2) (-3) (-2) (-1) \\ \\ = & \textcolor{springgreen}{\boldsymbol{12}} \\ \\ \end{aligned}