Prove that `|{:(betagamma,betagamma'+beta'gamma,beta'gamma'),(gammaalpha,gammaalpha'+gamma'alpha,gamma'alpha'),(alphabeta,alphabeta'+alpha'beta,alpha'beta'):}|`
`=(alphabeta'-alpha'beta)(betagamma'-beta'gamma)(gammaalpha'-gamma'alpha)`.

Options

A. `(alphabeta'-alpha'beta)(betagamma'-beta'gamma)(gammaalpha'-gamma'alpha)`
B. `(alphaalpha'-betabeta')(betabeta'-gammagamma')(gammagamma'-alphaalpha')`
C. `(alphabeta'+alpha'beta)(betagamma'+beta'gamma)(gammaalpha'+gamma'alpha)`
D. None of these

Answer

A

Explanation

`(a)` `(1)/(alphabetagamma)|{:(alphabetagamma,alphabetagamma'+alphabeta'gamma,alphabeta'gamma'),(alphabetaalpha,betagammaalpha'+betagamma'alpha,betagamma'alpha'),(alphabetagamma,alphabeta'gamma+alpha'betagamma,alpha'beta'gamma):}|`
Multiplying `R_(1)` by `alpha`, `R_(2)` by `beta` and `R_(3)` by `gamma` and dividing the determinant by `alphabetagamma` we have
`=(1)/(alphabetagamma)*alphabetagamma|{:(1,alphabetagamma'+alphabeta'gamma,alphabeta'gamma'),(1,betagammaalpha'+betaalphagamma',betagamma'alpha'),(1,gammaalphabeta'+gammaalpha'beta,gammaalpha'beta'):}|`
by `{:(R_(3)toR_(3)-R_(1)),(R_(2)toR_(2)-R_(1)):}`
`=|{:(1,alphabetagamma'+alphabeta'gamma,alphabeta'gamma'),(0,gamma(alpha'beta-alphabeta'),gamma'(alpha'beta-alphabeta')),(0,beta(alpha'gamma-alphagamma'),beta'(alpha'gamma-alphagamma')):}|`
`=(alpha'beta-alphabeta')(alpha'gamma-alphagamma')|{:(1,alphabetagamma',+,alphabeta'gamma,alphabeta'gamma'),(0,,gamma,,gamma'),(0,,beta,,beta'):}|`
`=(alpha'beta-alphabeta')(alpha'gamma-alphagamma')(gammabeta'-betagamma')`