Let `A=[[0,2q,r] , [p,q,-r] , [p,-q,r]]` If `A A^T=I_3` then `|p|=`

Options

A. `1/sqrt2`
B. `1/sqrt6`
C. `1/sqrt3`
D. `1/sqrt5`

Answer

A

Explanation

`A=[(0,2q,r),(p,q,-r),(p,-q,r)]`
`A A^(T)=I_3 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 [(4q^2_r^2,2q^2-r^2,-2q^2+r^2),(2q^2-r^2,p^2+q^2+r^2,p^2-q^2-r^2),(-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) ]`
`rArr {:(p^2+q^2+r^2=1),(p^2-q^2-r^2=0):}}rArr 2p^2=1 rArr |p|=1/sqrt2`