<s>[INST] Different values of $A^n$ using Cayley-Hamilton Theorem And Direct Multiplication Let's say there's a matrix A
$$
A = \begin{pmatrix}
    3 & -4\\
    1 & -1\\
    \end{pmatrix}
$$
Now I want to find $A^n$
I tried the following two methods but get different answers. Any solution ?
By Direct Multiplication:
$$
A^2= \begin{pmatrix}
    5 & -8\\
    2 & -3\\
    \end{pmatrix}
$$
$$
A^3= \begin{pmatrix}
    7 & -12\\
    3 & -5\\
    \end{pmatrix}
$$
$$
A^4= \begin{pmatrix}
    9 & -16\\
    4 & -7\\
    \end{pmatrix}
$$
Observing the pattern
$$
A^n= \begin{pmatrix}
    2n+1 & -4n\\
    n & 1-2n\\
    \end{pmatrix}
$$
Using Cayley-Hamilton Theorem:
The characteristic equation of the matrix A is
$$
(\lambda-1)^2=0
$$
Using Cayley-Hamilton Theorem:
$$
(A-I)^2=0
$$
$$\implies A=I$$
$$\implies A^n=I$$
Now clearly
$$
I \neq \begin{pmatrix}
    2n+1 & -4n\\
    n & 1-2n\\
    \end{pmatrix}
$$
Which is correct ??
 [/INST] $(A-I)^{2}=0$ does not imply that $A=I$.
[For example, $M=\left[\begin{array}{llll}0 & 1  \\ 0 & 0   \end{array}\right]$  is a matrix whose square is $0$. Bur $M$ itself is not $0$].
C-H Theorem gives $A^{2}=2A-I$. A simple induction argument gives $A^{n}=nA-(n-1)I$ for all $n$.
</s>