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I've got a matrix $A$ with a size of $(n \times n)$, that can be described like this:

$$A=\begin{bmatrix} 1 & 2 & 3 & \cdots & n \\ -1 & 0 & 3 & \cdots & n \\ -1 & -2 & 0 & \cdots & n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -1 & -2 & -3 & \cdots & 0 \\ \end{bmatrix} $$

How would I go about and find the determinant of this matrix (I thought about describing it as a sum but I don't know how to create an equation for any element $a_(ij_)$)?

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closed as off-topic by Lord Shark the Unknown, Leucippus, Claude Leibovici, user91500, Shailesh Sep 15 '17 at 9:29

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Hint: try adding the first row to each of the other rows. That doesn't change the determinant and gives you an upper-triangular matrix.

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  • $\begingroup$ "Upper-triangular matrix" is something everyone needs to know about, really makes things simpler. Thanks a lot! $\endgroup$ – Avamander Sep 14 '17 at 21:23

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