Question regarding 3 x 3 matrices If $A$ is a $3 \times 3$ matrix with real elements and $\det(A)=1$, then are these affirmations equivalent:
$$
\det(A^2-A+I_3)=0 \leftrightarrow \det(A+I_3)=6 \text{ and } \det(A-I_3)=0?
$$
 A: The characteristic polynomial $c_A(s)=det(sI-A)$ 
$det(A)=c_A(0)=1$ meaning the constant in the third degree polynomial is 1
$det(A^2 - A + I) = 0$ implies that $p_A(s)=s^2-s+1$ is a polynomial that takes a subspace $A$ acts on to $0$, hence $p_A(s)|m_a(s)|c_A(s)$ where $m_A(s)$ is the minimal polynomial of $A$.
$c_A(s)$ has degree three. $p_A(s)=s^2-s+1|c_A(s)$ therefore $c_A(s)=p_A(s) \times (s-a), a \in \mathbb R $. But $p_A(s)(s-a)$ will give constant term of $c_A(s)$ as $a$. But we already know that the constant term of $c_A(s)$ is $1$.
Therefore a=$-1$ and consequently $c_A(s)=(s^2-s+1)(s-1)$
consequently $det(A-1)=c_A(1)=0$ and $det(A+1)=c_A(-1)=6$
A: $\Leftarrow)$ From $\det (A-I)=0$ we know that $1$ is an eigenvalue. Let $x,y$ be the two others (possibly complex, and counting multiplicities). From $\det A=1$ we know that $xy=1$. And $6=\det(A+I)=2(x+1)(y+1)$, so 
$$
3=xy+x+y+1=2+x+y,
$$
so $x+y=1$. We get a system of two equations on $x,y$, namely
$$
x+y=1,\ \ xy=1.
$$
That is, $x(1-x)=1$, or $x^2-1x+1=0$. Note that $y$ satisfies the same equation. In any case $A^2-A+I$ has zero as eigenvalue, so $\det(A^2-A+I)=0$. 
$\Rightarrow)$ We consider the three eigenvalues of $A$, $x,y,z$; they satisfy $xyz=1$ by hypothesis. Since $\det(A^2-A+I)=0$, one of them, say $y$, satisfies $y^2-y+1=0$. This implies that $y$ is non-real, and its conjugate is also an eigenvalue (since $A$ has real entries). So $y=\frac12+i\frac{\sqrt3}2$, $z=\frac12-i\frac{\sqrt3}2$, and 
$$
x=\frac1{yz}=\frac1{\frac14+\frac34}=1.
$$
Now $\det(A-I)=0$ (since $1$ is an eigenvalue) and
$$
\det(A+I)=2(y+1)(z+1)=2\left(\frac32+i\frac{\sqrt3}2\right)\left(\frac32-i\frac{\sqrt3}2\right)=2\,\left(\frac94+\frac34\right)=6.
$$
