$$

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:

- Add the
**i**-th with the**(i-1)**-th; - Add the last row to all, getting two determinants, one of which is
**0**(because of the**(n-1)**-st column); - 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?