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Let A= [[a,b,c],[b,c,a],[c,a,b]] is an o...

Let `A= [[a,b,c],[b,c,a],[c,a,b]]` is an orthogonal matrix and `abc = lambda (lt0).`
The equation whose roots are `a, b, c, ` is

A

`x^(3)- 2x^(2) + lambda = 0`

B

`x^(3) -lambda x^(2) + lambda x + lambda = 0 `

C

`x^(3) - 2 x^(2) + 2 lambda x + lambda = 0 `

D

`x^(3) pm x^(2) - lambda = 0 `

Text Solution

Verified by Experts

The correct Answer is:
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)
Equation whose roots are a, b,c is
`x^(3) - (a+b+c) x^(2) + (ab+ bc+ca) x - abc = 0`
`rArr x^(3) - (pm1) = 0 -lambda = 0`
`therefore x^(3) pm x^(2) - lambda = 0`
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