I am asked to find the following determinant $$D = \begin{vmatrix} 1 & 2 & \cdots & n \\\ n+1 & n+2 & \cdots & 2n \\\ \vdots & \vdots & \vdots & \vdots \\\ n(n-1)+1 & n(n-1)+2 & \cdots & n^2 \end{vmatrix} \ $$ where $n\in\mathbb N$. Any help. Thank you.
Evaluating this determinant
2
$\begingroup$
linear-algebra
determinant
1 Answers
8
Step 1: Substract line 1 to line 2 and to line 3.
Step 2: Line 3 is now twice line 2, hence D=0.
Step 3: Check what happens when there is no line 3, that is, when n=1 or n=2.
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0As you suggested I got $D=1$ when $n=1$, $D=-2$ when $n=2$. Right? – 2012-08-26
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0@NancyR Yes that is correct. – 2012-08-26