<s>[INST] 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?
 [/INST] 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 $$\det(A_{n+1}) = (-1)^{n}(1)(n+1)$$
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