InterviewSolution
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InterviewSolution
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. For what values of `lambda` and `mu` the system of equations `x+y+z=6, x+2y+3z=10, x+2y+lambdaz=mu` has (i) Unique solution (ii) No solution (iii) Infinite number of solutions |
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Answer» we can write the above system of equations in the matrix form `[(1,1,1),(1,2,3),(1,2,lambda)][(,x),(,y),(,z)]=[(,6),(,10),(,mu)]` `rArr AX=B` where A`A=[(1,1,1),(1,2,3),(1,2,lambda)],X=[(,x),(,y),(,z)]andB=[(,6),(,10),(,mu)]` `therefore` augmented matrix `C=[A:B]=[(1,1,1,vdots,6),(1,2,3,vdots,10),(1,2,lambda,vdots,mu)]` Appying `R_(2) to R_(2) and R_(3) to R_(3)-R_(1),` we get `C=[(1,1,1,vdots,6),(0,1,2,vdots,4),(0,1,lambda-1,vdots,mu-6)]` Applying `R_(3) to R_(3)-R_(2),` we get (I) No solution `p(A)!=p(c)` `i.e.," " lambda-3!=0 and mu-10!= 0` (ii) A unique solution `p(A) = p(c) =3` `ie.," " lambda-3and mu in R` `therefore" " lambda!=3 and mu in R` (iii) Infinite number of solutions `p(A)=p(c)(angle3)` `ie.," " lambda-3=0 and mu -10=0` `therefore " " lambda=3 and mu =10` |
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