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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1please test it on some concrete matrix,for example third order,fourth order – dato datuashvili Jan 26 '14 at 07:28
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And then prove it by induction :) – user76568 Jan 26 '14 at 07:36
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i like this word induction :D – dato datuashvili Jan 26 '14 at 07:37
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@ghugni is there any question or problem was solved? – dato datuashvili Jan 26 '14 at 08:30
3 Answers
1
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 $(-1)^{n}*n!$
dato datuashvili
- 9,194
1
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:
$$|A|=(-1)^nn!$$
user76568
- 4,542
0
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.
Danny obama
- 78