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This question already has an answer here:

Let $M_n= \begin{bmatrix} 1 & 2 & 3 & ... & n\\ 2 & 2 & 3 & ... & n\\ 3 & 3 & 3 & ... & n\\ \vdots & \vdots & \vdots & \ddots & n\\ n & n & n & n & n \end{bmatrix}$

What's $\det(M_n)$ ?

Seems like it is one of these exercises virtually impossible to do without the one trick, which is why I ask this question.

Computing the first few determinants yields:

$\det(M_1)=1$

$\det(M_2)=-2$

$\det(M_3)=3$

so if I were to make a wild guess, I would say $\det(M_n)=n(-1)^{n+1}$

The induction process didn't work too well for me because I can't find the determinant once a row and a column have been added to pad the matrix to dimension $n+1$.

I tried to use the well-known formula for the determinant by cutting the inside in four parts ($M_n$ in the upper-left, $(n+1)$ in the lower-right, and the $(n+1)$ row and column, but at the end I must compute the determinant of a sum of matrices which didn't help...

I don't know the name of this specific matrix, so maybe this is a repost, but if this is the case, I can't find the original one

I'd be grateful if someone could provide some hints.

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marked as duplicate by Jean Marie, G Tony Jacobs, Community Sep 6 '17 at 17:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ That's a neat question. May I ask where you found it? $\endgroup$ – Theo Bendit Sep 6 '17 at 16:49
  • $\begingroup$ @TheoBendit Thanks, I found this on 4chan. oi67.tinypic.com/dxeqsm.jpg $\endgroup$ – Evariste Sep 6 '17 at 16:52
  • $\begingroup$ That... was unexpected. Cool problem though! $\endgroup$ – Theo Bendit Sep 6 '17 at 16:54
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Replace row 2 by row 2 minus row 1.

Replace row 3 by row 3 minus row 2.

Replace row 4 by row 4 minus row 3 etc.

You get $$\pmatrix{1&2&3&\cdots&n\\ 1&0&0&\cdots&0\\ 1&1&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&\cdots1&0}.$$

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Multiply $\frac{n}{n-1}$ to the $(n-1)$-th row and subtract it to the $n$th row. Then the matrix will be in form of $\pmatrix{M_{n-1}&n\mathbf{1}\\ \mathbf{0}&-\frac{n}{n-1}}$

where $\mathbf{1}=(1, 1, \cdots, 1)_{n-1}^T, \mathbf{0}=(0, 0, \cdots, 0)_{n-1}$. Thus inductively we have $\det(M_n)=-\frac{n}{n-1}\det(M_{n-1})$.

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Subtract the second row from the first row, then subtract the third row from the second row and so on until you subtract the last row from the penultimate row. You will be left with a lower triangular matrix with $-1$'s on the leading diagonal, apart from the last element that will be $n$. Hence the deteminant will be $\color{blue}{(-1)^{n+1}n}$.

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