<s>[INST] If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
    \dfrac{5}{2} & \dfrac{3}{2}\\
    -\dfrac{3}{2} & -\dfrac{1}{2} 
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it comes out to be something like this,
$$A^2 = \begin{bmatrix}
    4 & 3\\
    -3 & -2
\end{bmatrix}$$
$$ A^3 = \begin{bmatrix}
    \dfrac{11}{2} & \dfrac{9}{2}\\
    -\dfrac{9}{2} & -\dfrac{7}{2} 
\end{bmatrix}$$
$$A^4 = \begin{bmatrix}
    7 & 6\\
    -6 & -5
\end{bmatrix}$$
so $$ A^n = \begin{bmatrix}
    \dfrac{3n}{2}+1 & \dfrac{3n}{2}\\
    -\dfrac{3n}{2} & -\dfrac{3n}{2}+1 
\end{bmatrix} \tag{1}\label{eq1}$$
thus after substituting $n=2022$  we get $ A^{2022} = \begin{bmatrix}
    3034 & 3033\\
    -3033 & -3032
\end{bmatrix}$
$$\\$$
Is there is any other way to find the value of $A^{2022}$ without finding out the pattern and then general equation \eqref{eq1} like I did above?
 [/INST] Another option is to see that $A = I + B$, where
$B = \begin{pmatrix}
\frac 3 2 & \frac 3 2 \\
-\frac 3 2 & -\frac 3 2
\end{pmatrix}$
and $B^2 = 0$.
So $A^n = I + nB$.
But that's a post-hoc explanation: it is not obvious to remark that when looking at the matrix - although such a simplification can be expected, in an exercise like this one, especially if it comes from a math competition where speed matters.
</s>