if determinant `|{:( cos (0 + phi),,-sin (0+phi),,cos 2phi),(sin 0,,cos 0,,sin phi),(-cos 0,,sin0,,cos phi):}|` is

Options

A. independent of `theta` for all `lambda in R`
B. independent of `theta` and `alpha` when `lambda=1`
C. independent of `theta` and `alpha` when `lambda=-1`
D. independent of `lambda` for all `theta`

Answer

A, C

Explanation

`(a,c)` `|{:(cos(theta+alpha),-sin(theta+alpha),cos2alpha),(sintheta,costheta,sinalpha),(-costheta,sintheta,lambdacosalpha):}|`
`=(1)/(sinalphacosalpha)|{:(cos(theta+alpha),-sin(theta+alpha),cos2alpha),(sinthetasinalpha,costhetasinalpha,sin^(2)alpha),(-costhetacosalpha,sinthetacosalpha,lambdacos^(2)alpha):}|`
[Multiplying `R_(2)` and `R_(3)` by `sin alpha ` and `cos alpha`, respectively]
`=(1)/(sinalphacosalpha)xx|{:(0,0,cos2alpha+sin^(2)alpha+lambdacos^(2)alpha),(sinthetasinalpha,costhetasinalpha,sin^(2)alpha),(-costhetacosalpha,sinthetacosalpha,lambdacos^(2)alpha):}|`
[Applying `R_(1)toR_(1)+R_(2)+R_(3)`]
`=(cos2alpha+sin^(2)alpha+lambdacos^(2)alpha)/(sinalpha*cosalpha)|{:(sinthetasinalpha,costhetasinalpha),(-costhetacosalpha,sintheta cosalpha):}|`
`=(cos^(2)alpha+lambdacos^(2)alpha)|{:(sintheta,costheta),(-costheta,sintheta):}|=(1+lambda)cos^(2)alpha`
Therefore, the given determinants is independent of `theta` for all real values of `lambda`.
Also , `lambda=-1`, then it is independent of `theta` and `alpha`.