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.

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 $4^5\cdot \frac{7}{2} = 4^4 \cdot 14$
