If $a_{ij}=\max(i,j)$, calculate the determinant of $A$ 
If $A$ is an $n \times n$ real matrix and 
$$a_{ij}=\max(i,j)$$ 
for $i,j = 1,2,\dots,n$, calculate the determinant of $A$.

So, we know that 
$$A=\begin{pmatrix}
1 & 2 & 3 & \dots & n\\ 
2 & 2 & 3 & \dots & n\\ 
3 & 3 & 3 & \dots & n\\ 
\vdots & \vdots & \vdots & \ddots  & \vdots\\ 
 n& n & n & \dots & n
\end{pmatrix}$$
but what do I do after?

Apply row operations $R_{j} \leftarrow R_{j} - R_{1}, \, \forall j \in \left \{2, ...n\right \}$ on $A$ to get the following matrix:
$A_1 = \begin{pmatrix}
 1& 2 & 3 & ... & n-1 & n\\ 
 1&  0& 0 & ... &  0& 0\\ 
 2&  1& 0 & ... &  0& 0\\ 
 ...&  ...& ... & ... &  ...& ...\\ 
 n-2& n-3 & n-4 &  ...&  0& 0\\ 
 n-1&  n-2&  n-3&  ...&  1& 0
\end{pmatrix}$
Now, apply the following row operations in the order of $j = 2,3,...,n-1$,
$R_{k} \leftarrow R_{k} - (k-1)R_{j}, \, \forall k \in \left \{j+1, j+2, ..., n\right \}$
to get the following matrix:
$A_2 = \begin{pmatrix}
 1& 2 & 3 & ... & n-1 & n\\ 
 1&  0& 0 & ... &  0& 0\\ 
 0&  1& 0 & ... &  0& 0\\ 
 ...&  ...& ... & ... &  ...& ...\\ 
 0& 0 & 0 &  ...&  0& 0\\ 
 0&  0&  0&  ...&  1& 0
\end{pmatrix}$
Now apply the following row operations $R_1 \leftarrow (j-1)R_j, \, \forall j \in {2,3,...,n}$ to get the following matrix:
$A_3 = \begin{pmatrix}
 0& 0 & 0 & ... & 0 & n\\ 
 1&  0& 0 & ... &  0& 0\\ 
 0&  1& 0 & ... &  0& 0\\ 
 ...&  ...& ... & ... &  ...& ...\\ 
 0& 0 & 0 &  ...&  0& 0\\ 
 0&  0&  0&  ...&  1& 0
\end{pmatrix}$
Note that $A_3$ has a value of $n$ at $(1,n)$th cell, all $1$s on the first lower off-diagonal and all other elements are $0$s. Now, compute the determinant along the first row to get:
$|A_3| = (-1)^{n+1}n$
Since, all the above row operations do not change the value of the determinant, we obtain,
$|A| = |A_1| = |A_2| = |A_3| = (-1)^{n+1}$
