Calculate the determinant of $A-5I$ Question
Let $ A = 
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lambda = 0$ is $3$, i.e. $\lambda_1 = \lambda_2 = \lambda_3 = 0.$
Now trace($A$) = $\lambda_4 + \lambda_5 = 1+6+13+19+25 = 65$
Then det($A-5I$) = $(\lambda_1-5)(\lambda_2-5)(\lambda_3-5)(\lambda_4-5)(\lambda_5-5)=(-5)^3(\lambda_4\lambda_5 - 5 \times 65 + 25)$
We need to calculate the value of $\lambda_4 \lambda_5$, which includes sum of lots of determinant of $2 \times 2$ matrices.
Is there any fast way to calculate the determinant?

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
$$
\det(A - 5I) = \prod_{k=1}^5 (\lambda_k - 5) = \frac 14(55^2 - 25 \cdot 209) \cdot (-5)^3 = 68750
$$

Another strategy to calculate this determinant:
Note that $A$ has a rank factorization $A = UV$, where
$$
U = \pmatrix{0&1\\5&1\\10&1\\15&1\\20&1}, \quad 
V = \pmatrix{1 & 1 & 1 & 1 & 1\\
1 & 2 & 3 & 4 & 5}.
$$
The Sylvester determinant identity then yields
$$
\det(UV - 5 I) = 5^{3} \det(VU - 5 I) = (-5)^3 \det \pmatrix{45 & 5\\200 & 10} = 68750.
$$

If you want to find the eigenvalues:
Note that whenever matrices $U,V$ are such that $UV$ and $VU$ are well defined products, $UV$ and $VU$ will have the same non-zero eigenvalues. Thus, the non-zero eigenvalues of $A = UV$ are equal to the non-zero eigenvalues of
$$
B = VU = \pmatrix{1 & 1 & 1 & 1 & 1\\
1 & 2 & 3 & 4 & 5}
\pmatrix{0&1\\5&1\\10&1\\15&1\\20&1} = 
\pmatrix{50 & 5\\200 & 15}.
$$
The eigenvalues of this matrix (both of which are non-zero) are equal to the solutions of the characteristic equation
$$
\lambda^2 - \operatorname{tr}(B) \lambda + \det(B) = 0 \implies\\
\lambda^2 - 65 \lambda - 250 = 0  \implies \\
\lambda = \frac 12\left(65 \pm 5 \sqrt{209}\right).
$$
