I stumbled upon this question online:
$$A=\begin{bmatrix}5&6&&\cdots&6 \\6&5&6&\cdots&6 \\&&\ddots&& \\6&\cdots&6&5&6 \\6&\cdots&&6&5 \end{bmatrix}\in \mathbb R^{82\times 82}$$
I also found a way to solve this in Chegg:
Is there a better way to solve this problem maybe by reducing it to upper triangular form?
a matrix where every entry is $1$ has eigenvalues $0,0,0,0,..., 0,n.$ For all entries $6,$ we get $0,0,0,0,..., 0,6n.$ Then subtract $I$
For Sylvester's Law of Inertia...the outcome confirms that $n-1$ eigenvalues are zero, then a single positive eigenvalue. We can check that the nonzero eigenvalue is $n$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array} \right) $$
$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{array} \right) $$
$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{array} \right) $$
Render the product of the eigenvalues.
Let $M$ be the matrix and $I$ be the identity matrix. Then $M+I$ which consists of all $6$'s has rank $1$, meaning the eigenvalue $-1$ has multiplicity $82-1=81$. There is one more eigenvalue with multiplicity $1$, obtained by observing that all eigenvalue including multiplicity must sum to the trace of $M$. The trace is $410$, the other $81$ eigenvalues (all $-1$) add up to $-81$, therefore the last eigenvalue is $410-(81×(-1))=491$.
Then the determinant is $(-1)^{81}×491=\color{blue}{-491}$.
One can generalise: there is a general formula for the $n{\times}n$ determinant $$D_n(a,b)=\begin{vmatrix}b & a & a &\dots &a\\ a & b& a&\dots&a \\\vdots&\vdots&\vdots&\dots&\vdots\\a&a&a&\dots &b \end{vmatrix}.$$
Consider the $n{\times}n$ matrix $A$ with all coefficients equal to $a$. Its rank is $1$, so $\dim(\ker A)=n-1$, and its trace is $na$, so its characteristic polynomial is $$\chi_A(X)=\det(XI_n-A)=X^{n-1}(X-na).$$ On en déduit que \begin{align} D_n(a,b)&=\det\bigl((b-a)I_n+A\bigr)=(-1)^n\det\bigl((a-b)I_n-A\bigr)\\ &= (-1)^n(a-b)^{n-1}\bigl((a-b)-na\bigr)=\color{red}{(b-a)^{n-1}\bigl(b+(n-1)a\bigr)}. \end{align}
In the last example given, $a=6$, $b=5$, $n=82$, so that we indeed obtain $\;(-1)^{81}(5+81\cdot 6)=-491$.
