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 If the system of linear equations x4y+7z=g,  3y5z=h,  2x+5y9z=k  is consistent, then 

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a
2g+h+k=0
b
g+2h+k=0
c
g+h+k=0
d
g+h+2k=0

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detailed solution

Correct option is A

here,D=1−4703−5−25−9=1(−27+25)+4(0−10)+7(0+6)  expanding along R1=−2−40+42=0The system of linear equations have infinite many solutions. [   system is consistent and does not have unique solutions as D=0] ⇒    D1=D2=D3=0 Now, D1=0⇒g−47h3−5k5−9=0⇒g(−27+25)+4(−9h+5k)+7(5h−3k)=0⇒−2g−36h+20k+35h−21k=0⇒−2g−h−k=0⇒2g+h+k=0


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