Show that $\det(A)=2^{p}$

Question

We have a $(n×n)$-matrix $A$ with complex entries such that $\,A^{2}=3A-2I$.$~$ Show that there exists $p\in\{0,1,2,...,n\}$ such that $\det(A)=2^{p}$. I don't know if my proof is good. I took the polynomial $G(x)=x^{2}-3x+2=(x-1)(x-2)$. So $A$ is a solution for $G(x)$. The $A$'s minimal polynomial divides $G(x)$ so the minimal polynomial can be $(x-1)$ or $(x-2)$. So we will get that $\det(A)=2^{0}$ or $\det(A)=2^{n}$.

Answer 1

If initial relationship

$$A^2-3A+2I=0,\tag{1}$$

is applied to an eigenvector $V$ (with associated eigenvalue $\lambda$), one gets

$$(\lambda^2-3\lambda+2)V=0,$$

giving a quadratic equation with roots $\lambda = 2$ and $\lambda=1$, these eigenvalues being possibly multiple eigenvalues (with order $p$ and $q$ resp.).

Therefore, as the determinant of a matrix is the product of its eigenvalues :

$$\det(A)=2^p1^q=2^p$$

Answer 2

As a hint: try this \begin{align} A^2 &= 3A-2I \\ A^2-I &= 3A-3I \end{align} in the left side you can replace $I$ by $I^2$ in the right side factor $3$ then simplify. Can you take over?

!c$$A^2-I^2 = 3I(A-I)\\(A-I)(A+I)=3I(A-I)\\ \to \\A-I=0\\ OR \\ A+I=3I \to A=2I$$