Calculate the determinant of $A-5I$

To determine $\det(A - 5I)$, it suffices to compute the eigenvalues of $A$. To that end, we note that $A$ has rank $2$, which means that it has at most $2$ non-zero eigenvalues.

Let $\lambda_1,\lambda_2$ denote these eigenvalues. We note that $$ \lambda_1 + \lambda_2 = \operatorname{tr}(A) = 65\\ \lambda_1^2 + \lambda_2^2 = \operatorname{tr}(A^2) = 4725 $$ Solving these equations for $\lambda_{1},\lambda_2$ yields $$ \lambda_{1,2} = \frac 12\left(65 \pm 5 \sqrt{209}\right) $$ We thereby compute $$ \det(A - 5I) = \prod_{k=1}^5 (\lambda_k - 5) = \frac 14(55^2 - 25 \cdot 209) \cdot (-5)^3 = 68750 $$

Row-reducing the matrix $A-5I$ is quicker. $$ \begin{gather} \begin{pmatrix} -4 & 2 & 3 & 4 & 5 \\ 6 & 2 & 8 & 9 & 10 \\ 11 & 12 & 8 & 14 & 15 \\ 16 & 17 & 18 & 14 & 20 \\ 21 & 22 & 23 & 24 & 20 \end{pmatrix} \longrightarrow \begin{pmatrix} -4 & 2 & 3 & 4 & 5 \\ 14 & -2 & 2 & 1 & 0 \\ 23 & 6 & -1 & 2 & 0 \\ 32 & 9 & 6 & -2 & 0 \\ 37 & 14 & 11 & 8 & 0 \end{pmatrix} \\ \longrightarrow \begin{pmatrix} -4 & 2 & 3 & 4 & 5 \\ 14 & -2 & 2 & 1 & 0 \\ -5 & 10 & -5 & 0 & 0 \\ 60 & 5 & 10 & 0 & 0 \\ -75 & 30 & -5 & 0 & 0 \end{pmatrix} \longrightarrow \begin{pmatrix} -4 & 2 & 3 & 4 & 5 \\ 14 & -2 & 2 & 1 & 0 \\ -5 & 10 & -5 & 0 & 0 \\ 50 & 25 & 0 & 0 & 0 \\ -70 & 20 & 0 & 0 & 0 \end{pmatrix} \\ \longrightarrow \begin{pmatrix} -4 & 2 & 3 & 4 & 5 \\ 14 & -2 & 2 & 1 & 0 \\ -5 & 10 & -5 & 0 & 0 \\ 50 & 25 & 0 & 0 & 0 \\ -110 & 0 & 0 & 0 & 0 \end{pmatrix} \end{gather} $$ This takes not more than $30$ multiplications in all. The determinant is clearly $$5\cdot (-1) \cdot (-5) \cdot (-25) \cdot (-110) = 68750.$$ I chose to row-reduce the matrix in this fashion to make use of the fact that every entry in the last column is a multiple of $5$. It made the calculations easier when doing it by hand.

Let the standard basis of $\mathbb R^5$ be denoted $\{e_1,e_2,\dotsc,e_5\}$. Let $\vec l$ denote the leftmost column of $A$, and let $\vec 1$ be the column vector of all $1$s. We have that for all $i \in \{1,\dotsc,5\}$: $$\begin{align}&Ae_i = \vec l+(i-1)\vec1 \\&\vec l = \sum_{i=1}^5 ie_i \\&\vec 1 = \sum_{i=1}^5 e_i\end{align}$$ Since every vector in the image (or range, if you call it that) of $A$ is a linear combination of $\vec 1$ and $\vec l$, it follows that the rank of $A$ is $2$. From which it follows that $3$ of $A$'s eigenvalues are $0$.

We now proceed to find the remaining eigenvalues. We focus on the subspace of $\mathbb R^5$ spanned by $\vec 1$ and $\vec l$. We see that $$A \vec 1 = 10\vec 1 + 5\vec l \\ A \vec l= 160\vec 1 + 55\vec l$$ Restrict $A$ to the subspace $\operatorname{span}\{\vec 1, \vec l\}$, and change to the basis $\{\vec 1, \vec l \}$. You get:$$B=\begin{bmatrix}10 & 160 \\ 5 & 55 \end{bmatrix}$$ Any eigenvalue of $B$ is an eigenvalue of $A$. The eigenvalues of $B$ are $λ_{12} = \frac52 (13 \pm \sqrt{209}) $.

We've now found all of $A$'s eigenvalues.

Finally, the determinant of $A$ is $\prod_{i=1}^5(\lambda_i - 5) = \frac52(11 + \sqrt{209})\frac52(11 - \sqrt{209})(-5)(-5)(-5)=68750$.