Let A be a 3 x 3 matrix with characteristic polynomial $det(A - \lambda I) = -\lambda^3 + 3\lambda - 2$. Find det(A) and tr(A) Question: Let A be a 3 x 3 matrix with characteristic polynomial $det(A - \lambda I) = -\lambda^3 + 3\lambda - 2$. Find det(A) and tr(A)
I have no idea how to start this question. I know that the properties for det(A) and tr(A) is
If A has eigenvalues $\lambda_1,\lambda_2,\cdots,\lambda_n$, then det(A) = $\lambda_1,\cdots,\lambda_n$ and tr(A) = $\lambda_1 + \lambda_2 + \cdots + \lambda_n$
But I have no idea how to use this definition to answer this question. 
$\lambda^3 + 3\lambda - 2 = -(\lambda-1)^2 (\lambda + 2)$
$det(A) = -2 \cdot 1 = -2 ? $
$tr(A) = -2 + 1 = -1 ?$ 
 A: This answer is just to give you a little more insight for this problem. 
The general form of characteristic polynomial of $n\times n$ matrix $A$ can be written as follows
Char(A) = $\lambda^n - trace(A) \lambda^{n-1} + $terms of lower degree. Adding to this, the constant term of the characteristic polynomial of the matrix $A$ is $(-1)^n det(A)$.
In your problem the characteristic polynomial of the matrix $A$ can be written as   $\lambda^3 - 3\lambda + 2 = 0$
From the above argument it is clear that $Trace(A) = 0$ (Since coefficient of $\lambda^2 = 0$) and $det(A) = (-1)^3 \times (constant~ term) = (-1)^3 \times 2 = -2$.
This result can be useful in the computation of trace and determinant of the matrix without computing its eigenvalues. As in many cases you may find it difficult to factorize the  polynomial in $\lambda$ to compute the eigenvalues. Also, there is a result that 
For higher degrees poynomial, no general formula exists (or more precisely, no formula in terms of addition, subtraction, multiplication, division, arbitrary constants and $n-th $roots). This result is proved in Galois theory and is known as the Abel-Ruffini theorem. For more details you can click here.
A: The characteristics equation of $A$ is given by: 
\begin{align*}
\left|A-\lambda I\right|&=0\\
\implies -\lambda^3 + 3\lambda - 2&=0\hspace{25pt}\cdots(i)\\
\implies(\lambda-1)^2(\lambda+2)&=0
\end{align*}
So the eigen values are given by: $\lambda_1=1,\ \lambda_2=1,\ \lambda_3=-2$.
Now $Tr(A)=$ Sum of eigen values $=\lambda_1+\lambda_2+\lambda_3=0$, and
$det(A)=$ Product of eigen values $=\lambda_1\cdot\lambda_2\cdot\lambda_3=-2$.
Or, you can apply directly Vita's formula in (i) which gives:
Sum of roots of polynomial $=Tr(A)=-\dfrac{0}{-1}=0$, and
Product of roots of polynomial $=det(A)=-\dfrac{-2}{-1}=-2$.
