Problem with powers of matrices. Suppose that the matrix $A$ is as follows:
$$A=\begin{bmatrix}
    -3&4\\
     2&3\end{bmatrix}$$
We need to prove that $A^{2n + 1}=A$.
The way I tackled this problem is as follows:


*

*If $A^{2n + 1} =A$,  then $A^{2n}$ must be the same as the identity matrix $I$.

*Thus $(A^2)^n$ must be the same as $I$.

*By calculation $A^2=17I$.

*Thus the statement can't be proved.


I'm not sure if I'm correct. I believe we have to make use of Cayley Hamilton's Theorem and diagonalization to solve the problem but I can't seem to wrap my head around it. Any help will be appreciated.
 A: If one considers the matrix
$$
A=\left[\begin{array}{cc}-3&4s_1\cr2s_2&3\end{array}\right],
$$
where $s_1,s_2 \in \{-1,1\}$ with $s_1s_2=-1$, then the characteristic polynomial $p_A$ of $A$ is given by
$$
p_A(\lambda)=\lambda^2-\text{trace}(A)\lambda+\det(A)=\lambda^2-1.
$$
It follows from Cayley Hamilton's Theorem that $p_A(A)=0$, i.e.
$$
A^2=I.
$$
Hence $A^{2n}=I$ and $A^{2n+1}=A$ for every $n \in \mathbb{Z}$.
A: EDIT: As @Hagen states in the comment on the question, that perhaps one of the signs is wrong in the matrix $A$. This working follows if that is true. If not, then this answer can be discarded.
The characteristic polynomial of $A$ is
$$f(\lambda)=\lambda^2-1=(\lambda+1)(\lambda-1).$$
You can show that this matrix is diagonisable (you ahould check this), so that we can write $A=VDV^{-1}$ and $g(A)=Vg(D)V^{-1}$, where here $g(x)=x^{2n+1}$.
Since $D$ is a diagonal matrix with entries of $1$ and $-1$, then $D^{2n+1}=D$ with $n\in\mathbb{Z}$.
Thus,
$$A^{2n+1}=VD^{2n+1}V^{-1}=VDV^{-1}=A$$
when $n$ is an integer.
A: The entries of the matrix as given are false. Change the following $(A)_{2,1}$=-2 instead of 2 and it works. Solve simply by induction on $n$, it is very easy then. 
A: Show $A^3=A$ by direct computation.
Then proceed by induction on $n$:
$$A^{2n+1}= A^{2(n-1)} A^3 = A^{2(n-1)} A = A^{2(n-1)+1} = A$$
No need for Cayley-Hamilton.
