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?
 A: Let $d_n$ be the determinant of the $n\times n$ matrix
We can also write it as a recurrence
By expanding on the last row (or column) we observe that all but the minors of last two columns have linear dependent columns, so we have:
$d_n=-\frac{n^2}{n-1}d_{n-1}+nd_{n-1}=-\frac{n}{n-1}d_{n-1}$
Coupled with $d_1=1$ we get $d_n=(-1)^{n-1}n$
A: Hint to a more artsy proof:
Whenever $\mathcal{A}$ is a logical statement, we shall write $\left[\mathcal{A}\right]$ for the integer $\begin{cases} 1, & \text{ if } \mathcal{A} \text{ is true}; \\ 0, & \text{ if } \mathcal{A} \text{ is false} \end{cases}$. (This is called the Iverson bracket notation.)
Let $B$ be the $n\times n$-matrix whose $\left(i,j\right)$-th entry is $n \left[j=n\right] - \left[i \leq j < n\right]$. Here is how $B$ looks like:
$$
B = \left( \begin{array}{ccccc} -1 & -1 & \cdots & -1 & n \\ 0 & -1 & \cdots & -1 & n \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & -1 & n \\ 0 & 0 & \cdots & 0 & n \end{array} \right) .
$$
The matrix $B$ is upper-triangular, and so $\det B = n \left(-1\right)^{n-1}$.
Let $C$ be the $n \times n$-matrix whose $\left(i,j\right)$-th entry is $\left[i \geq j\right]$. Here is how $C$ looks like:
$$
C = \left(\begin{array}{ccccc} 1 & 0 & \cdots & 0 & 0 \\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & \cdots & 1 & 0 \\ 1 & 1 & \cdots & 1 & 1 \end{array} \right) .
$$
The matrix $C$ is lower-triangular, and so $\det C = 1$.
Now, check that $A = BC$, and conclude.
A: Use row operation to transform it to
$$
\begin{pmatrix}
1 & 2 & 3 & ... &n \\ 
1 & 0 & 0 & ... &0 \\ 
3 & 3 & 3 & ... & n\\ 
 ...&...  &...  &...  &... \\ 
 n& n & n & n & n
\end{pmatrix},
$$
then to (determinant is multiplied by $-1$)
$$
\begin{pmatrix}
1 & 0 & 0 & ... &0 \\ 
0 & 2 & 3 & ... &n \\ 
0 & 3 & 3 & ... & n\\ 
 ...&...  &...  &...  &... \\ 
 0& n & n & n & n
\end{pmatrix},
$$
and repeat the process.
Thus the determinant should be $(-1)^{n-1}n$.
Hope this helps.

It turns out this is just a reformulation of @G.Sassatelli's comment. Maybe I should delete it?
A: Hint: By induction on $n$. Subtract row $n-1$ from $n$, row $n-2$ from $n-3$ and so on. Then expand by cofactors along the last column.
A: You can also do $L_i-L_{i+1}$ which would give you a triangular matrix and the determinant is only the product of the coefficients of the diagonal
A: 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,
$|A| = |A_1| = |A_2| = |A_3| = (-1)^{n+1}$
A: Well you can prove this using induction too. The n = $1$ case is trivial. Let, $\det(A_{n}) = n(-1)^{n-1}$. We'll need to prove $$\det(A_{n+1}) = (-1)^{n}(n+1)$$Now $A_{n+1}$ looks like:-
$$A_{n+1}=\begin{pmatrix}A_{n}&P\\
        P^{T}&n+1
        \end{pmatrix}
$$
where $P^{T} = (n+1)(1,1,...1)^{T}$ $$\det(A_{n+1}) = \det(A_{n} - P\frac{1}{(n+1)}P^{T})(n+1)$$
$$\det(A_{n+1}) = \det(A_{n} - (n+1)J)(n+1)\cdots\cdots\cdots\cdots [1]$$
where J is a square matrix of only ones. Now,
$$-(A_{n} - (n+1)J)=
        \begin{pmatrix}
        n & n-1 & n-2 & \cdots &1 \\
        n-1 & n-1 & n-2 & \cdots & 1 \\
        n-2&n-2&n-2&\cdots &1\\
        \vdots & \vdots &\vdots &\ddots&\vdots\\ 
        1 & 1 & 1&\cdots& 1 \\
        \end{pmatrix}
$$
Inverse of such matrix is of the form:-
$$D = -(A_{n} - (n+1)J)^{-1}=
        \begin{pmatrix}
        1 & -1 & 0 & \cdots &0 \\
        -1 & 2 & -1 & \cdots & 0 \\
        0&-1&2&\cdots &0\\
        \vdots & \vdots &\vdots &\ddots&\vdots\\   
        0 & 0 & 0&\cdots& 2 \\
        \end{pmatrix}
$$
By expanding along row 1 we could break det(D) in terms of determinant of a simpler matrix B i.e.$$\det(D_{n}) = \det(B_{n-1})-\det(B_{n-2})$$ where, $$B =
        \begin{pmatrix}
        2 & -1 & 0 & \cdots &0 \\
        -1 & 2 & -1 & \cdots & 0 \\
        0&-1&2&\cdots &0\\
        \vdots & \vdots &\vdots &\ddots&\vdots\\   
        0 & 0 & 0&\cdots& 2 \\
        \end{pmatrix}
$$
B follows the recursion formulla $$\det(B_{n}) = 2\det(B_{n-1})-\det(B_{n-2})$$ Base case $\det(B_{1}) = 2$, which gives $\det(B_{n}) = n+1. $
Hence, For such matrices D we have $\det(D_{n}) = 1$ $\space$ $\forall n\geq2$
Now from [1] we have $$\det(A_{n+1}) = (-1)^{n}(1)(n+1)$$
