How can I find the determinant of this type of square matrix of $n\times n$ $$\begin{vmatrix} a&b&b&b& \ldots&b\\b&a&b&b&\ldots&b\\b&b&a&b&\ldots&b \\ \vdots&b&b&a&\ldots &b \\ b&b&b&\ldots &a&b \\b&b&b&b&\ldots &a \end{vmatrix}_{n\times n}=?$$
I have applied elementary operation: $C_1\to C_1+C_2+C_3+\ldots+C_n$ $$\begin{vmatrix} a+(n-1)b&b&b&b& \ldots&b\\a+(n-1)b&a&b&b&\ldots&b\\a+(n-1)b&b&a&b&\ldots&b \\ \vdots&b&b&a&\ldots &b \\ a+(n-1)b&b&b&\ldots &a&b \\a+(n-1)b&b&b&b&\ldots &a \end{vmatrix}_{n\times n}$$ Applied, $C_2\to C_2-C_3$, $C_3\to C_3-C_4$, $\ldots$, $C_{n-1}\to C_{n-1}-C_n$ $$=(a+(n-1)b)\begin{vmatrix} 1&0&0&0& \ldots&b\\1&a-b&0&0&\ldots&b\\1&0&a-b&0&\ldots&b \\ \vdots&0&0&a-b&\ldots &b \\ 1&0&0&\ldots &a-b&b \\1&0&0&\ldots &b-a&a \end{vmatrix}_{n\times n}$$ $$=(a+(n-1)b)(a-b)^{n-1}\begin{vmatrix} 1&0&0&0& \ldots&b\\1&1&0&0&\ldots&b\\1&0&1&0&\ldots&b \\ \vdots&0&0&1&\ldots &b \\ 1&0&0&\ldots &1&b \\1&0&0&\ldots &-1&a \end{vmatrix}_{n\times n}$$ I got stuck here. I can't find the way to solve this. However the answer is $(a+(n-1)b)(a-b)^{n-1}$.
your help will certainly be appreciated. thank you.
