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Probem

Suppose $\alpha \neq \beta $. Evaluate

$$D_n= \begin{vmatrix} x_1 & \alpha & \alpha &\cdots & \alpha &\alpha \\ \beta & x_2 & \alpha &\cdots & \alpha &\alpha \\ \beta & \beta & x_3 &\cdots & \alpha &\alpha \\ \vdots & \vdots & \vdots & &\vdots &\vdots \\ \beta & \beta &\beta&\cdots & x_{n-1} &\alpha \\ \beta & \beta &\beta&\cdots & \beta &x_n \\ \end{vmatrix}. $$

Solution

Notice that \begin{eqnarray*} D_n&&=\begin{vmatrix} x_1 & \alpha & \alpha &\cdots & \alpha &\alpha\\ \beta & x_2 & \alpha &\cdots & \alpha &\alpha\\ \beta & \beta & x_3 &\cdots & \alpha &\alpha\\ \vdots & \vdots & \vdots & &\vdots &\vdots\\ \beta & \beta &\beta&\cdots & x_{n-1} &\alpha\\ \beta & \beta &\beta&\cdots & \beta &x_n\\ \end{vmatrix}=\begin{vmatrix} x_1 & \alpha & \alpha &\cdots & \alpha &\alpha+0\\ \beta & x_2 & \alpha &\cdots & \alpha &\alpha+0\\ \beta & \beta & x_3 &\cdots & \alpha &\alpha+0\\ \vdots & \vdots & \vdots & &\vdots &\vdots\\ \beta & \beta &\beta&\cdots & x_{n-1} &\alpha+0\\ \beta & \beta &\beta&\cdots & \beta &\alpha+(x_n-\alpha)\\ \end{vmatrix}\\ &&=\begin{vmatrix} x_1 & \alpha & \alpha &\cdots & \alpha &\alpha\\ \beta & x_2 & \alpha &\cdots & \alpha &\alpha\\ \beta & \beta & x_3 &\cdots & \alpha &\alpha\\ \vdots & \vdots & \vdots & &\vdots &\vdots\\ \beta & \beta &\beta&\cdots & x_{n-1} &\alpha\\ \beta & \beta &\beta&\cdots & \beta &\alpha\\ \end{vmatrix}+\begin{vmatrix} x_1 & \alpha & \alpha &\cdots & \alpha &0\\ \beta & x_2 & \alpha &\cdots & \alpha &0\\ \beta & \beta & x_3 &\cdots & \alpha &0\\ \vdots & \vdots & \vdots & &\vdots &\vdots\\ \beta & \beta &\beta&\cdots & x_{n-1} &0\\ \beta & \beta &\beta&\cdots & \beta &x_n-\alpha\\ \end{vmatrix}\\ &&=\begin{vmatrix} x_1-\beta & \alpha-\beta & \alpha-\beta &\cdots & \alpha-\beta &0\\ 0 & x_2-\beta & \alpha-\beta &\cdots & \alpha-\beta &0\\ 0 & 0 & x_3-\alpha &\cdots &\alpha-\beta &0\\ \vdots & \vdots & \vdots & &\vdots &\vdots\\ 0 & 0 &0 &\cdots & x_{n-1}-\beta &0\\ \beta & \beta &\beta&\cdots & \beta &\alpha\\ \end{vmatrix}+(x_n-\alpha)D_{n-1}\\ &&=\alpha \prod_{i=1}^{n-1}(x_i-\beta)+(x_n-\alpha)D_{n-1}. \end{eqnarray*} Likewise,one can obtain $$D_n=\beta \prod_{i=1}^{n-1}(x_i-\alpha)+(x_n-\beta)D_{n-1}.$$ Since $\alpha \neq \beta$,it follows that $$D_n=\frac{1}{\alpha-\beta}\left[\alpha \prod_{i=1}^{n}(x_i-\beta)-\beta \prod_{i=1}^{n}(x_i-\alpha)\right].$$

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  • $\begingroup$ Looks good, I think! $\endgroup$ – StackTD May 3 at 14:16

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