Determinant

**james_bond** · March 28th 2009, 12:46 AM

$|A| = \begin{vmatrix} 1 & 2&3&\dots&n-1&n\\-1 & 0&3&\dots & n-1&n\\-1&-2&0&\dots &n-1 &n\\ \vdots&&&&&\\ -1& -2& -3&\dots & -(n-1) & 0 \end{vmatrix}$

Show that $|A|=n!$ .

**Opalg** · March 28th 2009, 01:10 AM

Originally Posted by james_bond

$|A| = \begin{vmatrix} 1 & 2&3&\dots&n-1&n\\-1 & 0&3&\dots & n-1&n\\-1&-2&0&\dots &n-1 &n\\ \vdots&&&&&\\ -1& -2& -3&\dots & -(n-1) & 0 \end{vmatrix}$

Show that $|A|=n!$ .

First, for $1\leqslant j\leqslant n$ you can take a factor j out of the j'th column, giving you $|A| = n!\begin{vmatrix} 1 & 1&1&\dots&1&1\\-1 & 0&1&\dots & 1&1\\-1&-1&0&\dots &1 &1\\ \vdots&&&&&\\ -1& -1& -1&\dots & -1 & 0 \end{vmatrix}$ . Now add the top row of that new matrix to each of the other rows and you'll get an upper-triangular matrix with 1s on the diagonal. So its determinant is 1.

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