I need to calculate the determinant of $A$, where $A$ is a $N$x$N$ matrix. With $A = (a_{ij})$ with $a_{ij} = i+j$
For a 1x1 matrix i found $det(A) = 2$ and same for a 2x2 $det(A) = 2$
But nof for a 3x3 I've founr $det(A) = 0$. So I got an inconsistency.
Anyone can help me further?
I will give here a matrix explanation.
Let us consider for example the case $4 \times 4$, which is in fact illustrative of the general case $n \times n$.
You can write $A$ as the sum:
$$\pmatrix{1&1&1&1\\2&2&2&2\\3&3&3&3\\4&4&4&4}+\pmatrix{1&2&3&4\\1&2&3&4\\1&2&3&4\\1&2&3&4}=\pmatrix{1\\2\\3\\4}\pmatrix{1&1&1&1}+\pmatrix{1\\1\\1\\1}\pmatrix{1&2&3&4}$$
which the sum of 2 rank-1 matrices, thus has rank $\leq 2$, and in fact effectively two.
Thus, due to the rank-nullity theorem, $\dim \ker A= 4 -2>0$ (and more generally $n-2$). Thus $\det(A)=0$.
It explains the "barrier" beyond $n=2$.
If $n\ge 3:$ $$\det A=\begin{vmatrix} 2 & 3 & \ldots & n+1\\ 3 &4 & \ldots & n+2 \\ 4 &5 & \ldots & n+3 \\ \vdots&&&\vdots \\ \end{vmatrix}\underbrace{=}_{R_2-R_1,R_3-R_2}\begin{vmatrix} 2 & 3 & \ldots & n+1\\ 1 &1 & \ldots & 1 \\ 1 &1 & \ldots & 1 \\ \vdots&&&\vdots \\ \end{vmatrix}=0.$$
Actually the value of $\operatorname {det}_{A_{1\times 1}} =2$ and $\operatorname {det}_{A_{2 \times 2}} =2\times 4 -3\times 3=-1$. After that the determinant is always zero.
That is because $R_3-R_1, R_4-R_1,\cdots R_n-1$ for a matrix $A _{n \times n}$ is always a multiple of $R_2-R_1$ where that entire row will be filled with ones. Hence, by the fundamental property, we have the determinant as zero. Hope it helps.