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 $$n \in N$$ for which $$P^n = 5I - 8P$$ is equal to ___.

We have $$P = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix}$$. First compute $$P^2 = P \cdot P = \begin{bmatrix} 4-5 & -2+3 \\ 10-15 & -5+9 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix}$$. Notice that $$P^2 = I - P$$, since $$-P + I = \begin{bmatrix} -2+1 & 1 \\ -5 & 3+1 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix}$$.

Using the recurrence $$P^2 = I - P$$, we compute successive powers. $$P^3 = P \cdot P^2 = P(I - P) = P - P^2 = P - (I - P) = 2P - I$$. Then $$P^4 = P \cdot P^3 = P(2P - I) = 2P^2 - P = 2(I-P) - P = 2I - 3P$$. Continuing: $$P^5 = P \cdot P^4 = P(2I-3P) = 2P - 3P^2 = 2P - 3(I-P) = 5P - 3I$$. Finally, $$P^6 = P \cdot P^5 = P(5P-3I) = 5P^2 - 3P = 5(I-P) - 3P = 5I - 8P$$.

Therefore $$P^n = 5I - 8P$$ when $$n = 6$$.