Let A be a matrix of order 3 such that A2=3A-2I where, I is an identity matrix of order 3. If A5=αA+β I, then α β is equal to

Question

Let A be a matrix of order 3 such that A2=3A-2I where, I is an identity matrix of order 3. If A5=αA+β I, then α β is equal to (A) 2025 (B) -2025 (C) -930 (D) -640

Tardigrade Admin · Accepted Answer

A3=A2(A)=(3 A - 2 I)A=3A2-2A =3(3 A - 2 I)-2A=7A-6I A4=A3A=(7 A - 6 I)A=7A2-6A=7(3 A - 2 I)-6A=15A-14I A5=A4A=(15 A - 14 I)A=15A2-14A=15(3 A - 2 I)-14A=31A-30I ⇒ α =31,β =-30

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

$2025$

$- 2025$

$- 930$

$- 640$

$A^{3} = A^{2} (A) = (3 A - 2 I) A = 3 A^{2} - 2 A$
$= 3 (3 A - 2 I) - 2 A = 7 A - 6 I$
$A^{4} = A^{3} A = (7 A - 6 I) A = 7 A^{2} - 6 A = 7 (3 A - 2 I) - 6 A = 15 A - 14 I$
$A^{5} = A^{4} A = (15 A - 14 I) A = 15 A^{2} - 14 A = 15 (3 A - 2 I) - 14 A = 31 A - 30 I$
$\Rightarrow α = 31, β = - 30$

Q. Let A be a matrix of order 3 such that A2=3A−2I where, I is an identity matrix of order 3. If A5=αA+βI, then αβ is equal to

2025

−2025

−930

−640