Question

Let $A$ be a matrix of order $3$ such that $A^{2}=3A-2I$ where, $I$ is an identity matrix of order $3.$ If $A^{5}=\alpha a+\beta I,$ then $\alpha \beta $ is equal to

• A. $2025$
• B. $-2025$
• C. $-930$
• D. $-640$

Answer 1

Answer: (C)

$A^{3}=A^{2}\left(A\right)=\left(3 A - 2 I\right)A=3A^{2}-2A=3\left(3 A - 2 I\right)-2A=7A-6I$
$A^{4}=A^{3}A=\left(7 A - 6 I\right)A=7A^{2}-6A=7\left(3 A - 2 I\right)-6A=15A-14I$
$A^{5}=A^{4}A=\left(15 A - 14 I\right)A=15A^{2}-14A=15\left(3 A - 2 I\right)-14A=31A-30I$
$\Rightarrow \alpha =31,\beta =-30$