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 ??
$(A-I)^{2}=0$ 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 $n$.