Using B2=I−B repeatedly:
B3=B⋅B2=B(I−B)=B−B2=B−(I−B)=2B−I.
B4=B⋅B3=B(2B−I)=2B2−B=2(I−B)−B=2I−3B.
B5=B⋅B4=2B−3B2=2B−3(I−B)=5B−3I.
B6=B⋅B5=5B2−3B=5(I−B)−3B=5I−8B.
So n=6.
If a square matrix B satisfies B^2 = I - B and B^n = 5I - 8B, then the value of n is :
CUET UG 2023 — Mathematics Applied-Mathematics
2023mcqhard
If a square matrix B satisfies B2=I−B and Bn=5I−8B, then the value of n is :
Official previous-year question
Held on 7 Jun 2023 · Verified 13 Jul 2026.
Options
- A
4
- B
6
- C
8
- D
10
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