<s>[INST] Determinant of a matrix with diagonal entries $a$ and off-diagonal entries $b$ 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.

 [/INST] 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
$$\det(A)=(a+(n-1)b)\begin{vmatrix}
      1 & 1 & 1 & \cdots & 1 \\
      0 & a-b & 0 &\cdots & 0 \\
     0 & 0 & a-b &\cdots & 0 \\
      \vdots & \vdots & \vdots  & \ddots & \vdots \\
      0 & 0 & 0 & \ldots & a-b \\
    \end{vmatrix}=(a+(n-1)b)(a-b)^{n-1}.$$
</s>