# linear algebra – Find the determinant of nth order

$$begin{vmatrix} 2 & 2 & … & 2 & 2 & 1 \ 2 & 2 & … & 2 & 2 & 2 \ 2 & 2 & … & 3 & 2 & 2 \ … & … & … & … & … & … \ 2 & n-1 & … & 2 & 2 & 2 \ n & 2 & … & 2 & 2 & 2 end{vmatrix}$$
I got this in my linear algebra homework. In the task, it is required to find the determinant of a matrix by the method of representing the sum of determinants. By that I mean this property of determinants:
$$begin{vmatrix} a & b+e \ c & d+f \ end{vmatrix} = begin{vmatrix} a & b \ c & d \ end{vmatrix} + begin{vmatrix} a & e \ c & f \ end{vmatrix}$$

What I have tried:

1. Add the i-th with the (i-1)-th;
2. Add the last row to all, getting two determinants, one of which is 0 (because of the (n-1)-st column);
3. Tried to get n on the diagonal, this is what I got:
$$begin{vmatrix} 2 & 3 & … & n-1 & n & n \ 2 & 3 & … & n-1 & n & n+1 \ 2 & 3 & … & n & n & n+1 \ … & … & … & … & … & … \ 2 & n & … & n-1 & n & n+1 \ n & 3 & … & n-1 & n & n+1 end{vmatrix}$$
The closest one to mine from StackExchange was this one. But I didn’t manage to link these two determinants.

No matter how I transform it, nothing worked for me. Any ideas?