$\textrm{Let }r,s\in \mathbb{R}\textrm{ such that }r\neq s.\\ \textrm{Let }A\in M_{n\times n}\left ( \mathbb{R} \right )\textrm{ such that }A_{ii}=r\: \: \forall i\in \mathbb{N}\textrm{ such that }1\leq i\leq n\\ \textrm{ and }A_{ij}=s\: \: \forall i,j\in \mathbb{N}\textrm{ such that }1\leq i,j\leq n\textrm{ such that }i\neq j.\\ \textrm{What is the determinant of }A?$
I know that A is diagonalizable since it is real symmetric
Approach by computing the product of eigenvalues:
Note that $A = (r-s)I + sJ$ where $J$ is the all ones matrix. The eigenvalues of $J$ are $sn$ with multiplicity $1$, and $0$ with multiplicity $n-1$. You can then show that the eigenvalues of $A$ are then $r-s+ns = r + (n-1)s$ with multiplicity $1$, and $r-s$ with multiplicity $n-1$.
$A = s I + r B $
where $B$ has $1$'s on the off-diagonal entries, and a zero diagonal.
If $v_1 = [1, 1, ...., 1]^T $
Then $B v_1 = (n-1) v_1 $
and if $v_{k+1} = [1, -1, 0, 0, ...., 0]^T$ where the $1$ is the $k$-th position, for $k = 1, 2, ..., n-1 $, then
$ B v_{k+1} = - v_{k+1} $
Thus $B$ has the following eigenvalues: $(n-1), -1, -1, ...., -1$.
Therefore, $A$ has the following eigenvalues
$r + s (n-1) , r - s , r- s, ..., r-s$
Hence,
$|A| = (r - s)^{(n-1)} (r + s(n-1) ) $
As an explicit example, if
$A = \begin{bmatrix} 5 && 2 && 2 \\ 2 && 5 && 2 \\ 2 && 2 && 5 \end{bmatrix}$
Then evaluating directly we get,
$|A| = 5(21) - 2 (6) + 2 (-6) = 105 - 24 = 81 $
Using the formula,
$|A| = (5 - 2)^2 (5 + 2 (2) ) = 81 $
