Let $A=(a_{ij})$ be a $n \times n$ matrix where $a_{ij} = \max(i,j)$. Find the determinant of $A$. 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?
 A: Hint: For $2\le k\le n$ substract the $k$-th row from the $k-1$-row.
A: 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 
$$
\text{Determinant} = (-1)^{n-1} n = - (-1)^n n$$
A: 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.
