Let α and β be real numbers. Consider a 3×3 matrix A such that A2=3A+αI. If A4=21A+βI, then [2023]
(1)
Given, A2=3A+αI and A4=21A+βI
Now, A2A2=(3A+αI)(3A+αI)
=9A2+6αIA+α2I=9(3A+αI)+6αIA+α2I
=27A+9αI+6αA+α2I=(27+6α)A+(9α+α2)I
So, 27+6α=21⇒α=-1
⇒9α+α2=β⇒9(-1)+(-1)2=β⇒β=-8