If matrix $A = \begin{pmatrix} 1 & -1 \\ 1 & -2 \end{pmatrix}$ satisfies $A^n = 5I - 8A$, then $n =$
Step-by-Step Solution
Key Concept: Use the minimal polynomial $A^2 + A - I = 0$ to express higher powers as linear combinations of $A$ and $I$.
Given $A = \begin{pmatrix} 1 & -1 \\ 1 & -2 \end{pmatrix}$ satisfies $A^2 + A - I = 0$, we compute successive powers: $A^2 = I - A$, $A^3 = A - A^2 = 2A - I$, $A^4 = 2A^2 - A = 2I - 3A$, $A^5 = 2A - 3A^2 = 5A - 3I$, $A^6 = 5A^2 - 3A = 5I - 8A$. The pattern follows the recurrence relation for powers of $A$, yielding $n = 6$.
Correct Answer: 3