So I'm studying for my course of linear algebra and the following problem was inside the book with exercises. "Given a matrix $A \in \mathbb{R}^{n \times n}$, where $a_{ij} = i + j$, calculate it's determinant".
The first thing I did was make a small visualization of what this sort of matrix would look like. I came up with this:
$$ A = \begin{pmatrix} 1 + 1 & 1 + 2 & ... & 1 + n\\ 2 + 1 & 2 + 2 & ... & 2 + n\\ ... & ... & & ...\\ n + 1 & n + 2 & ... & n + n \end{pmatrix}$$
I set of to set up an expression for the determinant using cofactor expansion (I don't know if this is the correct English terminology for it, the formula is given below, I also found on wikipedia this is called "Laplace Expansion"?) to the first row (thus making $j = 1$ in the expression):
$$ \sum^{n}_{i = 1}(-1)^{(i + 1)}a_{i1}det(M_{i1})$$
where $det(M_{i1})$ is the minor for the $i1$-th element. This is the point where I am stuck and could definitely use some help. I've tried writing $A$ in terms of summations to maybe see something in it, but this evolved into some very complex expression which I'm not even going to bother typing down (it's really, really long).
I have a feeling that I can bring the $a_{i1}$ out of the summation but I'm not sure how I would go on doing that.
Thanks in advance
The determinant is undefined if $m\neq n$, so consider matrices $A^{n\times n}$ with $a_{i,j}=i+j$.
For $n\ge3$, the determinant is $0$. The second row can be written as a linear combination of the first and third rows:
$$\begin{align} \begin{bmatrix}2&3&4&\cdots&1+n\end{bmatrix}^\top+\begin{bmatrix}4&5&6&\cdots&3+n\end{bmatrix}^\top&=\begin{bmatrix}6&8&10&\cdots&4+2n\end{bmatrix}^\top\\[1ex] &=2\begin{bmatrix}3&4&5&\cdots&2+n\end{bmatrix}^\top \end{align}$$
The cases of $n<3$ are trivial (but non-zero!).