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Today in exam I tried to evaluate this determinant but failed, only somehow "guessed" the answer I got here. Now in home I've managed to find something intuitive, just want to know whether the approach is correct, and is there more faster way exist. Given determinant $$\det\begin{vmatrix} 1 & 2 & 3 & ... & n-2 & n-1 & n\\ 2 & 3 & 4 & ... & n-1 & n & n\\ 3 & 4 & 5 & ... & n & n & n \\ . & . & . & . & . & . &. \\ n & n & n & ... & n & n & n \end{vmatrix}$$

First thing I did, was rearranging rows. I remember from another problem, where I used to evaluate determinant of matrix of this kind $\det\begin{vmatrix} 0 & 0 ... & 0 & 1\\ 0 & 0 ... & 1 & 0\\ 0 & 0 ... & 0 & 0\\ . & . & . & . \\ 1 & 0 ... & 0 & 0 \end{vmatrix}$ is $(-1)^{\frac{n(n-1)}{2}}*\det\begin{vmatrix} 1 & 0 ... & 0 & 0\\ 0 & 1 ... & 0 & 0\\ 0 & 0 ... & 0 & 0\\ . & . & . & . \\ 0 & 0 ... & 0 & 1 \end{vmatrix}=(-1)^{\frac{n(n-1)}{2}}$.
So it becomes $$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix} n & n & n & ... & n & n & n \\ . & . & . & . & . & . &. \\ 3 & 4 & 5 & ... & n & n & n \\ 2 & 3 & 4 & ... & n-1 & n & n\\ 1 & 2 & 3 & ... & n-2 & n-1 & n\\ \end{vmatrix}$$ And then I transposed it $$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix} n & n-1 & n-2 & ... & 3 & 2 & 1 \\ . & . & . & . & . & . &. \\ n & n & n & ... & n & n-1 & n-2 \\ n & n & n & ... & n & n & n-1\\ n & n & n & ... & n & n & n\\ \end{vmatrix}$$ and tried to subtract first row from all. $$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix} n & n-1 & n-2 & ... & 3 & 2 & 1 \\ 0 & 1 & 1 & ... & 1 & 1 & 1 \\ 0 & 1 & 2 & ... & 2 & 2 & 2\\ . & . & . & . & . & . &. \\ 0 & 1 & 2 & ... & n-3 & n-2 & n-1\\ \end{vmatrix}$$ next step is subtracting second row from others below. $$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix} n & n-1 & n-2 & ... & 3 & 2 & 1 \\ 0 & 1 & 1 & ... & 1 & 1 & 1 \\ 0 & 0 & 1 & ... & 1 & 1 & 1\\ . & . & . & . & . & . &. \\ 0 & 0 & 1 & ... & n-4 & n-3 & n-2\\ \end{vmatrix}$$ doing this for finite n we'll get $$(-1)^{\frac{n(n-1)}{2}}\det\begin{vmatrix} n & n-1 & n-2 & ... & 3 & 2 & 1 \\ 0 & 1 & 1 & ... & 1 & 1 & 1 \\ 0 & 0 & 1 & ... & 1 & 1 & 1\\ . & . & . & . & . & . &. \\ 0 & 0 & 0 & ... & 0 & 0 & 1\\ \end{vmatrix}$$ So my final solution is $n*(-1)^{\frac{n(n-1)}{2}}$. Do you see any mistakes? And maybe there's more easier approach? thanks.

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  • $\begingroup$ I think your approach is very correct. $\endgroup$ – Bman72 Jun 1 '15 at 14:32
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First we can subtract from each row the following row $$D= \begin{vmatrix} 1 & 2 & 3 & \ldots & n-2 & n-1 & n\\ 2 & 3 & 4 & \ldots & n-1 & n & n\\ 3 & 4 & 5 & \ldots & n & n & n \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ n & n & n & \ldots & n & n & n \end{vmatrix} = \begin{vmatrix} -1&-1 &-1 & \ldots &-1 &-1 & 0\\ -1&-1 &-1 & \ldots &-1 & 0 & 0\\ -1&-1 &-1 & \ldots & 0 & 0 & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ n & n & n & \ldots & n & n & n \end{vmatrix}$$ We can "pull out" $n$ from the last row, and then add the last row to all rows above it (to cancel out all $-1$'s) $$D=n \begin{vmatrix} -1&-1 &-1 & \ldots &-1 &-1 & 0\\ -1&-1 &-1 & \ldots &-1 & 0 & 0\\ -1&-1 &-1 & \ldots & 0 & 0 & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 1 & 1 & 1 & \ldots & 1 & 1 & 1 \end{vmatrix}= n \begin{vmatrix} 0& 0 & 0 & \ldots & 0 & 0 & 1\\ 0& 0 & 0 & \ldots & 0 & 1 & 1\\ 0& 0 & 0 & \ldots & 1 & 1 & 1\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 1 & 1 & 1 & \ldots & 1 & 1 & 1 \end{vmatrix} $$ Now we only need to subtract each row from the row following it to get $$ D= n \begin{vmatrix} 0& 0 & 0 & \ldots & 0 & 0 & 1\\ 0& 0 & 0 & \ldots & 0 & 1 & 1\\ 0& 0 & 0 & \ldots & 1 & 1 & 1\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 1 & 1 & 1 & \ldots & 1 & 1 & 1 \end{vmatrix}= n \begin{vmatrix} 0& 0 & 0 & \ldots & 0 & 0 & 1\\ 0& 0 & 0 & \ldots & 0 & 1 & 0\\ 0& 0 & 0 & \ldots & 1 & 0 & 0\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 1 & 0 & 0 & \ldots & 0 & 0 & 0 \end{vmatrix}=n(-1)^{\frac{n(n-1)}2}$$ (We got the matrix for which you already know the determinant. It can be obtained, for example, using several swaps of neighbouring rows.)

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  • $\begingroup$ beautiful :) thanks for your time sir. Have you looked at my approach for correctness? $\endgroup$ – shcolf Jun 1 '15 at 14:29
  • $\begingroup$ I did not spot any mistake there. (And I'd say that the two solutions are, to some extent, similar.) $\endgroup$ – Martin Sleziak Jun 1 '15 at 14:36

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