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Consider the system of equations x+y+...

Consider the system of equations
`x+y+z=6`
`x+2y+3z=10`
`x+2y+lambdaz =mu`
the system has unique solution if

A

`lamda=3, mu=10`

B

`lamda!=,mu=10`

C

`lamda!=3,mu!=10`

D

`lamda=3, mu!=10`

Text Solution

Verified by Experts

The correct Answer is:
D

Given `x+y+z=6,x+2y+3z=10`
and `x+2y+lamdaz=mu`
For no solution `A=|(1,1,1),(1,2,3),(1,2,lamda)|=0`
`implies1(2 lamda=6)-1(lamda-3)+1(2-2)=0`
`implieslamda-3=0implieslamda=3`
`[(2lamda-6,-(lamda-2),1),(-(lamda-3),lamda-1,-2),(0,1,1)][(6),(10),(mu)]!=[(0),(0),(0)]`
`implies[(12 lamda-36, -10lamda+20, +mu),(-6lamda+18xx, 10 lamda-10,-2mu),(0+10,+mu, )]!=[(0),(0),(0)]`
`implies[(2lamda-16+mu),(4 lamda+8-2mu),(10+mu)]!=[(0),(0),(0)]`
`implies2lamda-16+mu!=0`
`=2xx3-16+mu!=0`
`implies mu!=10`
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