Let `Delta_1=[[ap^2,2ap,1],[aq^2,2aq,1],[ar^2,2ar,1]]` and `Delta_2=[[apq,a(p+q),1],[aqr,a(q+r),1],[arp,a(r+p),1]]` then

Options

A. `Delta_(1)=Delta_(2)`
B. `Delta_(2)=2Delta_(1)`
C. `Delta_(1)=2Delta_(2)`
D. `Delta_(1)+2Delta_(2)=0`

Answer

D

Explanation

`(d)` `Delta_(1)=|{:(ap^(2),2ap,1),(aq^(2),2aq,1),(ar^(2),2ar,1):}|`
`=2a^(2)|{:(p^(2),p,1),(q^(2),q,1),(r^(2),r,1):}|`
`=-2a^(2)(p-q)(q-r)(r-p)`
`Delta_(2)=|{:(apq,a(p+q),1),(aqr,a(q+r),1),(arp,a(r+p),1):}|`
`=a^(2)|{:(pq,(p+q),1),(qr,(q+r),1),(rp,(r+p),1):}|`
`=a^(2)|{:(pq-qr,p-r,0),(qr-rp,q-p,0),(rp,(r+p),1):}|(R_(1)toR_(1)-R_(2),R_(2)toR_(2)-R_(3))`
`=a^(2)(p-q)(r-p)|{:(-q,-1,0),(-r,-1,0),(rp,(r+p),1):}|`
`=a^(2)(p-q)(r-p)|{:(-q+r,0,0),(-r,-1,0),(rp,(r+p),1):}|`
`=a^(2)(p-q)(r-p)(q-r)`
`impliesDelta_(1)+2Delta_(2)=0`