Matrix in $\mathbb{Z}_5$

Question

Let $A=\begin{bmatrix}3&2\\3&3\end{bmatrix} \in M_2(\mathbb{Z}_5).$ Then if I calculate $A^{105}$ like $105 \equiv 0 \pmod 5$ , $A^{105} = Id_2$ ? Thank you.

Answer 1

No.

But notice that $A^4 = I$, so $A^{105} = A^{4 \cdot 26 + 1} = (A^4)^{26} A = I^{26} A = A$.

Answer 2

As a general method to compute this kind of problem, we may make use of Hamilton-Cayley theorem, that $A=\begin{pmatrix}a& b\\c & d\end{pmatrix}$. Then $A^2-(a+d)A+(ad-bc)I=0$. Hence $$P(A)=(A^2-(a+d)A+(ad-bc)I)Q(A)+\alpha A+\beta I=\alpha A+\beta I.$$ So we may only compute the remainder for quotient for $P(A)$. As an example, let's take $P(A)=A^{105}$ in $\mathbb{Z}_5$. Thus $P(0)=0=\beta$. So $P(A)=\alpha A$. Note that $P(I)=I$, which means $\alpha=1$. We conclude that $A^{105}=A$.