<s>[INST] 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.
 [/INST] 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= $4^4×14$
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