Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?

*

*$4^4 \times 14$


*$4^3 \times 14$


*$4^2 \times 14$


*$4   \times 14$

I think we have
$$\det \left[\begin{array}{ll}6 & 2 \\ 2 & 6\end{array}\right] = 4 \times 8$$
$$\det \left[\begin{array}{lll}6 & 2 & 2 \\ 2 & 6 & 2 \\ 2 & 2 & 6\end{array}\right] = 4^{2} \times10$$
and for any $n$ we have $\det(A_n)= 4^{n-1} \times (6+2×(n-1))$ so "1" is true.

Your generalization, with an obvious but unstated definition of $A_n$, can be proved as follows. Define $E_n$ as the $n\times n$ matrix whose entry are all $1$s. You can show its only nonzero eigenvalue is $2n$, and the associated eigenspace is $1$-dimensional. By the rank-nullity theorem, the eigenvalue $0$ has multiplicity $n-1$. So $A_n=4I_n+2E_n$ has eigenvalues $4,\,2n+4$ with respective multiplicities $n-1,\,1$.
