I have a problem. To find and prove the formula for $det(A)$ of given matrix of size $n\times n$:
$a_{i,j}=2, i\neq j$, and $a_{i,i}=i$
So for example:
$det\begin{pmatrix} 1&2&2&2\\ 2&2&2&2\\ 2&2&3&2\\ 2&2&2&4\\ \end{pmatrix}=-4 $
What is the most straight-forward way to do so?
I was able to get the solution myself in the end, quite simple task to be honest when I got the idea. By doing the row operations, so by eliminating twos from 3rd to n-th row simply by subtracting 2nd row result matrix is:
$\begin{pmatrix} 1&2&2&2&...&2\\ 2&2&2&2&...&2\\ 0&0&1&0&...&0\\ 0&0&0&2&...&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&0&...&n-2\\ \end{pmatrix}$
Then eliminate 1st item on 2nd row, the result is now:
$\begin{pmatrix} 1&2&2&2&...&2\\ 0&-2&-2&-2&...&-2\\ 0&0&1&0&...&0\\ 0&0&0&2&...&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&0&...&n-2\\ \end{pmatrix}$
The matrix is now upper triangular, so we simply multiply the diagonal.
And result can be interpreted as:
$det(A)=1\space for\space n=1;\\det(A)=-2\cdot(n-2!)\space for\space n>1;n\in\mathbb{N}$