Generalized Method to find nth power of matrix in $P^n = 5I - 8P$ 
Question: Let $I$ be an identity matrix of order $2 \times 2$ and $P = \begin{bmatrix}
    2 & -1 \\
    5 & -3 \\
   \end{bmatrix}$. Then the value of $n ∈ N$ for which $P^n = 5I - 8P$ is equal to _______.
Answer: 6
Question Source: JEE Mains $18^{th}$ March Shift-2 2021

By characteristic Equation: $P^2-(\operatorname {Tr}(P))P+(\det (P))I=0$ we can find $n=6$ by hit and trial. i.e. multiplying equation by P gives
$P^3=(\operatorname {Tr}(P))P^2-(\det (P))P = (\operatorname {Tr}(P))[(\operatorname {Tr}(P))P+(\det(P))I]$
And going on solving for $P^6$. But is there a generalized method because it can't be done for high values of $n$ (eg:$100$)?

Here is a standard method which can be used both for calculating a high power of a $2 \times 2$ matrix and solving this problem. It can be extended to higher size matrix.
Let $X^2+aX+b$ be the characteristic polynomial. Then there exists some $a_n,b_n$ such that
$$
X^n=Q(X)(X^2+aX+b)+ a_nX+b_n \,...(1)
$$
$a_n,b_n$ can be calculated explicitely, by plugging in the equation the two roots $r_1,r_2$ of the characteristic polynomial.
Plugging $A$ in (1) you get
$$
A^n=a_nA+b_n \,.
$$
In case of a double root differentiate (1) with respect to $X$ gives ($\alpha$ is repeated eigenvalue)
$$nX^n=2Q(X)(X-\alpha) + Q'(X)(X-\alpha)^2 + a_n \,...(2)$$ put eigenvalue in (1) and (2)
In your particular problem $r_{1,2}=\frac{-1\pm \sqrt{5}}{2}$ and hence
$$
(\frac{-1\pm \sqrt{5}}{2})^n=a_n\frac{-1\pm \sqrt{5}}{2}+b_n 
$$
You need to figure out $n$ such that
$$
(\frac{-1\pm \sqrt{5}}{2})^n=-8\frac{-1\pm \sqrt{5}}{2}+5 
$$
