InterviewSolution
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InterviewSolution
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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 ` a^(2) b^(2) + b^(2) c^(2) + c^(2) a^(2)`, isA. `2lambda`B. `-2lambda`C. `lambda^(2)`D. `-lambda` |
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Answer» Correct Answer - B `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^(2) b^(2) + b^(2) a^(2) + c^(2) a^(2) = (ab + bc+ ca) ^(2) - 2abc (a + b + c)` `= 0- 2abc (pm 1) pm 2 lambda [ because abc = lambda ]` ` = - 2 lambda ` `[because lambda lt 0]` |
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