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Let `A= [[a,b,c],[b,c,a],[c,a,b]]` is an orthogonal matrix and `abc = lambda (lt0).` The value of ` a^(3) + b^(3)+c^(3)` isA. `lambda`B. `2lambda`C. `3lambda`D. None of these |
Answer» Correct Answer - D `becauseA` is an orthogonal matrix `therefore A A^(T) =I` `[[a,b,c],[b,c,a],[c,a,b]] [[a,b,c],[b,c,a],[c,a,b]] =1 [[1,0,0],[0,1,0],[0,0,1]]` `[[a^(2)+b^(2)+c^(2),ab + bc+ca,ab + bc+ ca],[ab + bc + ca,a^(2) +b^(2)+c^(2) , ab+ bc+ ca ],[ab+ bc+ca,ab+bc+ca,a^(2) + b^(2) + c^(2)]] =1 [[1,0,0],[0,1,0],[0,0,1]]` By equality of matrices, we get `a^(2) + b^(2) +c^(2) = 1 ` ...(i) `ab + bc + ca= 0` ...(ii) ` (a+b+c)^(2) + a^(2)= b^(2) +c^(2)+ 2 (ab + bc + ca)` `= 1 + 0 = 1` ` therefore a+ b + c = pm 1` ...(iii) ` because a^(3) + b^(3) +c^(3) - 3abc = (a+b+c) ` `(a^(2) + b^(2) +c^(2) - ab - bc - ca)` `rArr a^(3) + b^(3) + c^(3) - 3lambda = (pm 1) (1-0) ` [from Eqs.(i), (ii) and (iii) and abc` = lambda`] `rArr a^(3) + b^(3)+ c^(3) = 3lambda pm 1` |
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