Finding determinant from trace and $A^2-4A+3I =0$. Let $A$ and $B$ are two $3 \times 3$ matrices such that $A^2-4A+3I = B^2 - 4B+3I = 0$. Also it is given that $tr(A) = 7$ and $tr(B) = 5$. 
Can we conclude anything about determinant of $A$ and $B$?
 A: Each eigenvalue of $A,\,B$ is $1$ or $3$. So $A$ has eigenvalues $1,\,3,\,3$ of product $9$; similarly, $\det B=\det(1^23)=3$.
A: *

*Find the sum and the product of the two eigenvalues of $A$ from $A^2-4A+3I=0$ which have to satisfy $x^2-4x+3=0$ of which we can read it of the equation and conclude that the sum is $4$ and the product is $3$. 

*From the trace of $A$ which is $7$, we can determine that the last eigenvalue is $3$. Multiply $3$ with the product of the other two which is $3$, we conclude that the answer is $9$.

*Similarly for $B$.
Remark: If matrix $A$ has size $n$, and you are given degree $n-1$ polynomial that matrix $A$ satisifes, you do not need to solve for the root explicitly.
A: Let $v$ be eigenvector of $A$ with eigenvalue $\lambda$.
We have $Av= \lambda v$ .
Now multiply $(A^2-4A+3I)v=(\lambda^2 -4\lambda + 3I)v =0$.   
Vector $v$ is non-zero vector hence we have $\lambda^2 -4\lambda +3=0$.  
From this you have two eigenvalues.
Additionally the trace of $A$ is the sum of eigenvalues  ( which allows to obtain the third eigenvalue) and the determinant is the product of eigenvalues.
