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Let A = ((0,2q,r),(p,q,-r),(p,-q,r)). If...

Let `A = ((0,2q,r),(p,q,-r),(p,-q,r))`. If `A A^(T) = I_(3)`, then `|p|` is

A

`(1)/(sqrt(5))`

B

`(1)/(sqrt(2))`

C

`(1)/(sqrt(3))`

D

`(1)/(sqrt(6))`

Text Solution

Verified by Experts

The correct Answer is:
B
Given, `"AA"^(T) = I`
`rArr [{:(0, 2q, r), (p, q, -r), (p,-q, r):}][{:(0, p, p), (2q, q, -q), (r,-r, r):}] = [{:(1, 0, 0), (0, 1, 0), (0, 0, 1):}] `
`rArr [{:(0+4q^(2), 0+2q^(2)-r, 0-2q^(2) +r^(2)), (0+2q^(2) -r^(2), p^(2) + q^(2) + r^(2), p^(2) -q^(2)-r^(2)), (0-2q^(2) + r^(2), p^(2) - q^(2)-r^(2), p^(2) + q^(2) +r^(2)):}] = [{:(1, 0, 0), (0, 1, 0), (0, 0, 1):}]`
We know that, if two matrices are equal, then corresponding elements are also equal, so
`4q^(2) + r^(2) - 1 = p^(2) + q^(2) +r^(2), " " (i)`
`2q^(2) - r^(2) = 0 rArr r^(2) = 2q^(2) " " .....(ii)`
and ` p^(2)-q^(2)-r^(2) = 0 " "... (iii)`
Using Eqs. (ii) and (iii), we get
`p^(2) = 3q^(2) " " ...(iv)`
Using Eqs. (ii) and (iv) in Eq. (i), we get
`4q^(2) + 2q^(2) = 1`
`rArr 6q^(2) = 1`
`rArr 2p^(2) = 1 " " ["using Eq." (iv)]`
`p^(2) = (1)/(2) rArr |p| = (1)/(sqrt(2))`
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