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.
 A: As, dimension of nullspace of $(A-4I)$ is $4$. So, Geometric multiplicity of eigenvalue $4$ is $4$, as matrix $A$ is symmetric, so, $A$ must be diagonalizable, and hence, Algebraic multiplicity and Geometric multiplicity of eigenvalue $4$ is same, so, Algebraic multiplicity of eigenvalue $4$ is $4$.
And, as each row sum is $14$, so, $14$ is another eigenvalue of algebraic multiplicity $1$.
Det($A$)=multiplication of eigenvalues= $4^4×14$
A: $$\begin{vmatrix}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{vmatrix}=\begin{vmatrix}4 & 0 & 0 & 0 & -4 \\ 0 & 4 & 0 & 0 & -4 \\ 0 & 0 & 4 & 0 & -4 \\0 & 0 & 0 & 4 & -4 \\ 2 & 2 & 2 & 2 &\ \ 6\end{vmatrix}=\begin{vmatrix}4 & 0 & 0 & 0 & -4 \\ 0 & 4 & 0 & 0 & -4 \\ 0 & 0 & 4 & 0 & -4 \\0 & 0 & 0 & 4 & -4 \\ 0 & 0 & 0 & 0 & 14\end{vmatrix}$$
A: According to the Matrix Determinant Lemma, $\det(A+uv^T)= \det(A)(1+v^T A^{-1}u)$, hence
$$\det(A) = \det(4I + 2\cdot\mathbf 1 \mathbf 1^T) = \det(4I)(1+2\cdot \mathbf 1^T(\tfrac{1}{4}I)\mathbf 1) = 4^n (1 +\tfrac{1}{2}n) $$
For $n=5$, that's $4^5\cdot \frac{7}{2} = 4^4 \cdot 14$
A: 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 $n-1,\,1$.
