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If bc+qr=ca+rp=ab+pq=-1 and (abc,pqr!=0)...

If `bc+qr=ca+rp=ab+pq=-1` and `(abc,pqr!=0)` then `|[ap,a,p],[bq,b,q],[cr,c,r]|` is (A) 1 (B) 2 (C) 0 (D) 3

Text Solution

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we are given that
` 1+ bc+ qr=0`
`1+ca +pr=0`
`1+ab +pq =0`
The determinant in the question involves a column consisting the elements ap, bq and cr
So multipying (1),(2) and (3) by ap bq and cr respectively we get
`ap+ abcp + apqr =0`
`bq+ abcq+bpqr=0`
`cr+ abcr+cpqr=0`
Since abc and pqr occur in all the three equations putting abc= x pqr=y we get
`ap+ px+ ay=0`
`bq+ qx+by=0`
`cr+rx+cy=0`
This system must have a common solution (i.e., system must be consistent)
So ,`|{:(ap,,p,,a),(bq,,q,,b),(cr,,r,,c):}|=0`
`rArr |{:(ap,,a,,p),(bq,,b,,q),(cr,,c,,r):}|=0`
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