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Last updated at March 30, 2023 by Teachoo
Misc 13 If A =[■8("α" &"β" @"γ" &−"α" )] is such that A2 = I then A. 1 + α2 + βγ = 0 B. 1 – α2 + βγ = 0 C. 1 – α2 – βγ = 0 D. 1 + α 2 – βγ = 0 A = [■8("α" &"β" @"γ" &−"α" )] Given that A2 = I A . A = I [■8("α" &"β" @"γ" &−"α" )] [■8("α" &"β" @"γ" &−"α" )]= [■8(1&[email protected]&1)] [■8(𝛼. 𝛼+𝛽. 𝛾&𝛼. 𝛽+𝛽(−𝛼)@𝛾. 𝛼−𝛼. 𝛾&𝛾. 𝛽+(−𝛼)(−𝛼))]= [■8(1&[email protected]&1)] [■8(𝛼2+𝛽 𝛾&𝛼𝛽−𝛼𝛽@𝛼𝛾−𝛼𝛾&𝛾𝛽+𝛼2)]= [■8(1&[email protected]&1)] [■8(𝛼2+𝛽𝛾&[email protected]&𝛽𝛾+𝛼2)]= [■8(1&[email protected]&1)] Since the matrices are equal, corresponding element are equal ∴ 𝛼2+𝛽𝛾 = 1 0 = 1 – 𝛼2−𝛽𝛾 1 – 𝜶𝟐−𝜷𝜸 = 0 Hence, C is the correct answer