Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade 12
Question:
Matrix $A$ is such that $A^2 = 2A - I$, where $I$ is identity matrix, then for $n \geq 2, A^n =$
nA - (n-1)I
nA - I
2^(n-1)A - (n-1)I
2^(n-1)A - I
Step-by-Step Solution
Key Concept: Use the recurrence relation $A^2 = 2A - I$ to find a pattern in higher powers by repeated substitution.
For matrix $A$ satisfying $A^2 = 2A - I$, compute successive powers: $A^3 = 2A^2 - A = 4A - 2I - A = 3A - 2I$, $A^4 = 3A^2 - 2A = 6A - 3I - 2A = 4A - 3I$. The pattern shows $A^n = nA - (n-1)I$ for all $n \geq 1$.
Correct Answer: 1