Let $A=(a_{ij})$ be a $n \times n$ matrix where $a_{ij} = \max(i,j)$. Find the determinant of $A$. How to find the solution of this kind of problem?
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Similar problem with $\min$ instead of $\max$ can be found here: http://math.stackexchange.com/questions/392738/evaluation-of-a-specific-determinant/ – Martin Sleziak Jan 28 '14 at 09:27
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https://math.stackexchange.com/q/2013663/321264 – StubbornAtom Jan 26 '22 at 10:23
3 Answers
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Hint: For $2\le k\le n$ substract the $k$-th row from the $k-1$-row.
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Elaborating on what Not Me's answer.
If you subtract row 1 from row 2, row 2 from row 3 etc you get $$ \pmatrix{1&2&3&\cdots& n\\ 1 & 0 & 0 & \cdots & 0\\ 1 & 1 & 0 & \cdots & 0\\ 1 & 1 & 1 & \cdots & 0\\ \vdots & \vdots&\vdots & \ddots & 1 } $$ If you expand on the first row, you see that the determinant is $\pm n$ with $+$ if $n$ is odd and $-$ if $n$ is even. So the final answer is $$ \text{Determinant} = (-1)^{n-1} n = - (-1)^n n$$
user44197
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if $n=2$ $\det = -2$, and if $n=3$ $\det = -3$. Why do you have $1$ in the right down corner ? – Nikita Evseev May 22 '14 at 07:08
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Subtract the 1st row from the 2nd row we have the new 2nd row to be :
1 0 0 ... 0. Expand the determinant on this 2nd row we have it equal to:
-1*/A/ where A is the new minor of the form:
2 3 4 5 ... n
3 3 4 5.... n
4 4 4 5 ... n
Repeat step 1 by subtracting 1st row of A from 2nd row of A and expand along the 2nd row we get:
1*/B/ with B is the minor of A by deleting the 2nd row and 2nd column of A. Repeating this step on the way to the last step that we have: the final minor to be:
(n - 1) n
n n
this one has determinant = n(n - 1) - n*n = -n. So the final answer is: -n if n = even, and n if n = odd.
DeepSea
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I'm afraid it's not clear from your answer if the determinant is positive or negative. So... how do we decide? ;) – Jan 19 '14 at 05:34
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