suppose `D= |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}| ` and `Dprime= |{:(a_(1)+pb_(1),,b_(1)+qc_(1),,c_(1)+ra_(1)),(a_(2)+pb_(2),,b_(2)+qc_(2),,c_(2)+ra_(2)),(a_(3)+pb_(3),,b_(3)+qc_(3),,c_(3)+ra_(3)):}| `. Then

Options

A. `D'=D`
B. `D'=D(1-pqr)`
C. `D=D(1+p+q+r)`
D. `D'=D(1+pqr)`

Answer

D

Explanation

`D' = |{:(a_(1)+pb_(1),,b_(1)+qc_(1),,c_(1)+rq_(1)),(a_(2)+pb_(2),,b_(2)+qc_(2),,c_(2)+ra_(2)),(a_(3)+pb_(3),,b_(3)+qc_(3),,c_(3)+ra_(3)):}|`
`=|{:(a_(1),,b_(1)+qc_(1),,c_(1)+rq_(1)),(a_(2),,b_(2)+qc_(2),,c_(2)+ra_(2)),(a_(3),,b_(3)+qc_(3),,c_(3)+ra_(3)):}|+ |{:(pb_(1),,b_(1)+qc_(1),,c_(1)+rq_(1)),(pb_(2),,b_(2)+qc_(2),,c_(2)+ra_(2)),(pb_(3),,b_(3)+qc_(3),,c_(3)+ra_(3)):}|`
In the first determinant apply `C_(3) to C_(3) -rC_(1)` and then
`C_(2) to C_(2)-qC_(3)`
In the second determinant teke p common from `C_(1)` and then apply
`C_(2) to C_(2)-C_(1)` Then take q common from `C_(2) ` and apply
`C_(3) to C_(3) -C_(2)` Finally taking r common from `C_(3)` we have finally `D'=(1+pqr ) D`