Let `A= [[a,b,c],[b,c,a],[c,a,b]]` then find tranpose of A matrix

Answer

D

Explanation

`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`