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Let $A=(a_{ij})\in \mathbb{M}_n(\mathbb{R})$ be defined by

$$ a_{ij} = \begin{cases} i, & \text{if } i+j=n+1 \\ 0, & \text{ otherwise} \end{cases} $$ Compute $\det (A)$


After calculation I get that it may be $(-1)^{n-1}n!$. Am I right?

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    $\begingroup$ please test it on some concrete matrix,for example third order,fourth order $\endgroup$ Jan 26, 2014 at 7:28
  • $\begingroup$ And then prove it by induction :) $\endgroup$
    – user76568
    Jan 26, 2014 at 7:36
  • $\begingroup$ i like this word induction :D $\endgroup$ Jan 26, 2014 at 7:37
  • $\begingroup$ @ghugni is there any question or problem was solved? $\endgroup$ Jan 26, 2014 at 8:30

3 Answers 3

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for example let us take $3X3$ matrix,then it would be following matrix

a=[0 0 1;0 2 0;3 0 0]

a =

 0     0     1
 0     2     0
 3     0     0


det(a)

ans =

    -6

in your case if we compute $(-1)^{n-1}*n!=(-1)^2*n!=6$

maybe it is $(-1)^{n}*n!$

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These are matrices with only off-diagonal elements, with values being the row numbers. So the number of negative factors is the same for if it was a diagonal matrix: even for an even $n$, and odd for an odd $n$. So looks like the answer is:

$$|A|=(-1)^nn!$$

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  • $\begingroup$ i was thinking also like this,+1 $\endgroup$ Jan 26, 2014 at 7:30
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Just take $n=2$ then accordingly given conditions

$$A=\begin{bmatrix} 0 & 1\\ 2 & 0\\\end{bmatrix}$$ $Det(A)= -2=-2!$

& For $n=3$ $Det(A)= -6=-3!$

For $n=4$ $Det(A)=24=4!$

This suggests us the general formula for det. of such type of matrix is $$(-1)^{(n-1)(\frac{n}{2})}.n!$$

I think u can see it very easily Thanks.

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