Given the characteristic equation, how to find the determinant of a matrix Take a look at this question:

Find $\det(A)$ given that A has $p(\lambda)$ as its characteristic polynomial.
$$
p(\lambda) = \lambda^3 - 2\lambda^2 + \lambda + 5
$$

My first step is to notice the following:
$$
\det(\lambda I -A) = \lambda^3 - 2\lambda^2 + \lambda + 5
$$
But $\det(\lambda I-A) \neq \det(\lambda I) - \det(A)$. Any idea or hint to what to do with the aforementioned equation to compute $\det(A)$? Do I have to compute the eigenvalues and substitute them?
By the fundamental theorem of algebra, we see that
\begin{align}
p(\lambda) = (\lambda-z_1)(\lambda-z_2)(\lambda-z_3)
\end{align}
where $z_1, z_2, z_3$ are the roots which means
\begin{align}
p(0) = (-1)^3z_1z_2z_3 = -\det(A).
\end{align}
By Caley-Hamilton theorem a matrix satisfies its own charateristic equation so Here we have
$$A^3-2A^2+A+5I=0~~~~(1)$$
A $3 \times 3$ matrix $A$ satisfies a unique monic cubic:
$$A^3-Tarce(A)~ A^2+Trace(adj~ A)~ A-\det|A| I ~~~~(2)$$ By comparision
of (1) with (2), we get $\det|A|=-5.$

Also, for example $I^3-I=0 \implies \det|I|=1$