Q. Matrix A such that $A^{2} = 2A - I$, where $I$ is the identity matrix, Then for $n \geq 2,A^{n}$ is equal to
A) $2^{n - 1}A - (n - 1)I\$
B) $2^{n - 1}A - I$
C) $nA - (n - 1)I$
D) $nA - I$
Explanation
(c)
Given, $A^{2} = 2A - I$
Now, $A^{3} = A\left( A^{2} \right)$
$= A(2A - I)$
$= 2A^{2} - A$
$= 2(2A - I) - A$
$= 3A - 2I$
$A^{4} = A\left( A^{3} \right)$
$= A(3A - 2I)$
$= 3A^{2} - 2A$
$= 3(2A - I) - 2A$
$= 4A - 3I$
Following this, we can say $A^{n} = nA - (n - 1)I.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.