linear algebra - Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)

I have the following $n\times n$ matrix:

$$A=\begin{bmatrix} a & b & \ldots & b\\ b & a & \ldots & b\\ \vdots & \vdots & \ddots & \vdots\\ b & b & \ldots & a\end{bmatrix}$$

where $0 < b < a$.

I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$. This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out. Any help would be appreciated.

Answer

Add row 2 to row 1, add row 3 to row 1,..., add row $n$ to row 1, we get
$$\det(A)=\begin{vmatrix} a+(n-1)b & a+(n-1)b & a+(n-1)b & \cdots & a+(n-1)b \\ b & a & b &\cdots & b \\ b & b & a &\cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \ldots & a \\ \end{vmatrix}$$
$$=(a+(n-1)b)\begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ b & a & b &\cdots & b \\ b & b & a &\cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \ldots & a \\ \end{vmatrix}.$$
Now add $(-b)$ of row 1 to row 2, add $(-b)$ of row 1 to row 3,..., add $(-b)$ of row 1 to row $n$, we get
$$\det(A)=(a+(n-1)b)\begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & a-b & 0 &\cdots & 0 \\ 0 & 0 & a-b &\cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & a-b \\ \end{vmatrix}=(a+(n-1)b)(a-b)^{n-1}.$$