∣A∣=0αβ+6=0αβ=−6α+β=1⇒α=3,β=−2 A=[36−1−2]A2=[36−1−2][36−1−2]=[36−1−2]∴A2=AA=A2=A3=A4=A5(I+A)8=I+8C1 A7+8C2 A6+…..+8C8 A8=I+A(8C1+8C2+…..+8C8)=I+A(28−1)=[1001]+[7651530−255−510]
=[7661530−255−509]
Let A=[α6−1β],α>0, such that det(A)=0 and α+β=1. If I denotes 2×2 identity matrix, then the matrix (1+A)8 is:
Held on 2 Apr 2025 · Verified 6 Jul 2026.
[46−1−1]
[257514−64−127]
[10252024−511−1024]
[7661530−255−509]
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