Q. Matrix $A$ is such that $A^{2} = 2A - I$ where $I$ is the identity matrix, then for $n \geq 2,\ A^{n}$ is equal to
A) $n\ A - (n - 1)I$
B) $n\ A - I$
C) $2^{n - 1}A - (n - 1)I$
D) $2^{n - 1}A - I$
Explanation
(a)
As we have $A^{2} = 2A - I$
$\Longrightarrow \ \ A^{2}A = (2A - I)A = 2A^{2} - IA$
$\Longrightarrow \ \ \ A^{3} = 2(2A - I) - IA = 3A - 2I$
Similarly, $\ \ \ A^{4} = 4A - 3I$
$\ \ \ \ A^{5} = 5A - AI$
$\ \ \ \ A^{n} = 5A - 4I$
$\cdots$ $\cdots\ \ \ \ \cdots\ \ \ \cdots\ \ \cdots\ \ \cdots$
$\cdots$ $\cdots\ \ \ \cdots\ \ \ \cdots\ \ \ \ \cdots\ \ \ \cdots$
$A^{n} = nA - (n - 1)I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.