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I am given homework like this, calculate the Matrix $$ \begin{bmatrix}x+1 &x&x&...&x\\x&x+2&x&...&x\\x&x&x+3&...&x\\...&...&...&...&...\\x&x&x&...&x+n\end{bmatrix} $$

I tried to change it into triangular matrices but the best result I get is:

$$ \begin{bmatrix}x+1 &x&x&...&x\\-1&2&0&...&0\\-1&0&3&...&0\\...&...&...&...&...\\-1&0&0&...&n\end{bmatrix} $$

I wonder if I am on the right track since there seems to be no way to advanced.

Any hint given is appreciate.

Kurouku
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    You have made lots of progress. All those zeros should be heartening. Let $D(n)$ be the determinant of the order $n$ matrix. If you expand the last column, you get $D(n)=nD(n-1)+x(-1)^{n-1}(n-1)!$ – Ross Millikan Feb 25 '16 at 17:07
  • It seams that $\det A= (-1)^{n-1}a_nx+n!$ where $a_2=3 $,$a_3=11$, $a_4=50$, $a_5=274$, $a_6=1764$ are elements of the sequence A000254.(https://oeis.org/A000254). But i'm not able to do a proof :( – Emilio Novati Feb 25 '16 at 21:01
  • @RossMillikan I'm not able to find a pattern to that, how can I move on? – Kurouku Feb 26 '16 at 02:10
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    The correct recurrence, which supports what @EmilioNovati posted, is $D(n)=nD(n-1)+(n-1)!x$. I dropped a couple signs. – Ross Millikan Feb 26 '16 at 03:41

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