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I need to find a determinant of the matrix: $$ A = \begin{pmatrix} 1 & 2 & 3 & \cdot & \cdot & \cdot & n \\ x & 1 & 2 & 3 & \cdot & \cdot & n-1 \\ x & x & 1 & 2 & 3 & \cdot & n-2 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ x & x & \cdot & \cdot & x & 1 & 2 \\ x & x & \cdot & \cdot & \cdot & x & 1 \\ \end{pmatrix} $$ We know that $x \in R$ So far I managed to transform it to the form: $$ \begin{pmatrix} 1-x & 1 & 1 & \cdot & \cdot & \cdot & 1 \\ 0 & 1-x & 1 & 1 & \cdot & \cdot & 1 \\ 0 & 0 & 1-x & 1 & 1 & \cdot & 1 \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 0 & 0 & \cdot & \cdot & 0 & 1-x & 1 \\ x & x & \cdot & \cdot & \cdot & x & 1 \\ \end{pmatrix} $$ by the operations: (Let's say $r_i$ is the ith row) $$r_1 = r_1 - r_n,r_2 = r_2-r_n, r_3 = r_3 - r_n, ..., r_{n-1} = r_{n-1} - r_n$$ and then $$r_1 = r_1 - r_2, r_2 = r_2 - r_3, r_3 = r_3 - r_4,...,r_{n-2} = r_{n-2} - r_{n-1}$$ Unfortunately, I have no idea how to eliminate the last row. Any hints?

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  • $\begingroup$ It is actually easier to get the same effect by omitting the first sequence of operations altogether, and extending the second sequence by $r_{n-1}=r_{n-1}-r_n$ (the final operation of the removed first sequence). $\endgroup$ Commented Jan 14, 2015 at 23:04

2 Answers 2

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If $c_i$ is $i$th column of your second determinant, do $c_n= c_n-c_{n-1}$, $c_{n-1}=c_{n-1}-c_{n-2}$, ..., $c_2=c_2-c_1$ to get: $$\left|\begin{array}{ccccccc} 1-x & x & 0 & 0 & \cdots & 0 & 0\\ 0 & 1-x & x & 0 & \cdots & 0 & 0\\ 0 & 0 & 1-x & x & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & 1-x & x\\ x & 0 & 0 & 0 & \cdots & 0 & 1-x\\ \end{array}\right|$$

This determinant is obviously equal $(1-x)^n+(-1)^{n+1}x^n$ (expand it by the first column).

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  • $\begingroup$ Could you tell me which theory you're using? The first term is just a multiplication of entries on the diagonal (like in an upper-triangular matrix) but I don't know why there is this second term. $\endgroup$
    – wisniak
    Commented Jan 14, 2015 at 21:50
  • $\begingroup$ By expanding matrix by the first column you get the second term to be $(-1)^{n+1}xx^{n-1}$. It is $(-1)^{n+1}x$ since $x$ is in $(n,1)$ place, multiplied by the determinant of diagonal $(n-1)\times(n-1)$ matrix with $x$'s on diagonal. $\endgroup$
    – SMM
    Commented Jan 14, 2015 at 21:54
  • $\begingroup$ Your matrix is actually obtained as $JAJ^t$ from the original matrix $A$, where $J$ has diagonal entries $1$ and superdiagonal entries $-1$. $\endgroup$ Commented Jan 14, 2015 at 23:01
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Multiply the last row by $\frac{1-x}{x}$; this means that the determinant you want will be the determinant of the changed matrix times $-\frac{x}{x-1}$. Now subtract $r_1$ from $r_n$ leaving $$r_n = (0, -x, -x, -x, \cdots, -x, \frac{(x-1)^2 - x^2}{x}) $$ where I have intentionally written $$ \frac{1-x}{x} -1 = \frac{-2x+1}{x} = \frac{(x-1)^2 - x^2}{x} $$ Now we have 0 in the last row in columns 1 through 1. For each remaining column $j$ up to column $n-1$, multiply the last row by $\frac{x-1}{x}$ (giving another power of $\frac{x}{x-1}$ in the factor before the changed matrix), at which point you can eliminate the $(1-x)$ in column $j$ of the last row by adding the $j$-th row. When you do this, the last element of the $n$-th row changes to $$ \frac{(x-1)^{j}- x^{j}}{x^{j-1} } \frac{x-1}{x} - 1 = \frac{(x-1)^{j+1}-x^{j}(x-1) -x^{j}}{x^{j}} = \frac{(x-1)^{j+1}-x^{j+1}}{x^{j}} $$ and the process repeats for the mext $j$ In the end, the final term in $A_{nn}$ involves $$ \frac{(x-1)^n -x^n}{x^{n-1}}$$ and a lot of cancelation with the accumlated factors happens, leaving the answeer $$ (-1)^n \left( (x-1)^n - x^n \right)$$

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