Given, A2=3A+αI
∴A3=A2.A
⇒A3=3A2+αA
⇒A3=3(3A+αI)+αA
⇒A3=9A+αA+3αI
Now, A4=A3.A
A4=(9+α)A2+3αA
=(9+α)(3A+αI)+3αA
=A(27+6α)+α(9+α).......(1)
Given, A4=21A+βI.........(2)
On comparing equation (1) and equation (2),
⇒27+6α=21⇒α=−1
⇒β=α(9+α)=−8
JEE Main 2023 — Mathematics Algebra
Let α and β be real numbers. Consider a 3×3 matrix A such that A2=3A+αI. If A4=21A+βI, then
Held on 29 Jan 2023 · Verified 6 Jul 2026.
α=1
α=4
β=8
β=−8
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