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good morning I want to prove that $$D_n = \det \begin{pmatrix} x & 0 & \cdots & \cdots & 0 & 1 \\ -1 & x & 0 & \vdots & 0 & 2 \\ 0 &-1 & \ddots & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & x & 0 & n-2 \\ 0 & \cdots & 0 & - 1 & x & n-1 \\ 0 & \cdots & 0 & 0 & - 1 & x \\ \end{pmatrix} $$

$D_n = x^n + x^{n - 1} + \left( {n - 1} \right)x^{n - 2} + \cdots( {n - m + 1} )x^{n - m} + \cdots + 3x^2 + 2x + 1$

thank you in advance

sabaga
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1 Answers1

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Here's a proof that just uses row reduction. Let $R_i$ denote the $i^\text{th}$ row of your matrix. Replacing the first row by $R_1 \leftarrow xR_2 + x^2 R_3 + \cdots + x^{n-1} R_n$ results in the matrix $$ \begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & 1 + 2 x + 3 x^2 + \cdots + (n-1)x^{n-2} + x^n \\ -1 & x & 0 & \cdots & 0 & 2 \\ 0 & -1 & x & \cdots & 0 & 3 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & -1 & x \end{pmatrix} \, . $$

Using Laplace expansion along the first row, we have $$ \det(D_n) = (-1)^{n+1} (1 + 2 x + 3 x^2 + \cdots + (n-1)x^{n-2} + x^n) \det(M) \, , $$ where $M$ is the $(n-1) \times (n-1)$ matrix $$ \begin{pmatrix} -1 & x & 0 & \cdots & 0 \\ 0 & -1 & x & \cdots & 0 \\ \vdots & \vdots & \ddots & & \vdots \\ 0 & 0 & 0 & \cdots & -1 \end{pmatrix} \, . $$ Since $M$ is upper triangular, its determinant is the product of its diagonal entries, so $\det(M) = (-1)^{n-1}$.

Then \begin{align*} \det(D_n) &= (-1)^{n+1} (1 + 2 x + 3 x^2 + \cdots + (n-1)x^{n-2} + x^n) (-1)^{n-1} \\ &= (-1)^{2n} (1 + 2 x + 3 x^2 + \cdots + (n-1)x^{n-2} + x^n) \\ &= 1 + 2 x + 3 x^2 + \cdots + (n-1)x^{n-2} + x^n \, . \end{align*}

Viktor Vaughn
  • 19,278
  • After your first step (clearing the $x$ from the first row), you don't need to handle the other rows: Laplace expansion by the first row gives as only remaining determinant one of an upper triangular matrix with diagonal entries $-1$, in which the remaining entries $x$ have no effect. – Marc van Leeuwen Jun 07 '14 at 15:49
  • @MarcvanLeeuwen You're right, that does make things easier. I'll edit to that effect. – Viktor Vaughn Jun 07 '14 at 15:51
  • if $F(x) = x(x - 1)(x - 2) \cdots (x - n + 1)$;i want to calcul

    $\left| {\begin{array}{*{20}c} {F(0)} & {F(1)} & {F(2)} & { \cdots F(n)} \ {F(1)} & {F(2)} & {F(3)} & { \cdots F(n + 1)} \ \cdots & \cdots & \cdots & { \cdots \cdots } \ {F(n)} & {F(n + 1)} & {F(n + 2)} & { \cdots F(2n)} \ \end{array}} \right|$

     – sabaga Jun 08 '14 at 07:47
  • $[{\left( {F\left( n \right)} \right)^{n + 1}} = {\left( {n\left( {n - 1} \right)\left( {n - 2} \right)...\left( {n - n + 1} \right)} \right)^{n + 1}} = {\left( {n!} \right)^{n + 1}}]$ – sabaga Jun 08 '14 at 07:52
  • @sabaga If you have a new question, please submit it as a new question, not a comment. – Viktor Vaughn Jun 08 '14 at 14:42