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?
On
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 $I^3-I=0 \implies \det|I|=1$
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}