Consider the system of equations
`x+y+z=6`
`x+2y+3z=10`
`x+2y+lambdaz =mu`
The system has no solution if (a) `lambda ne 3` (b) `lambda =3, mu =10` (c) `lambda =3, mu ne 10` (d) none of these

Options

A. `lambda ne 3`
B. `lambda =3, mu =10`
C. `lambda =3, mu ne 10`
D. none of these

Answer

C

Explanation

`Delta = |{:(1,,1,,1),(1,,2,,3),(1,,2,,lambda):}|=2lambda +3+2 -2-lambda -6=lambda -3`
`Delta_(1) = |{:(6,,1,,1),(10,,2,,3),(mu,,2,,lambda):}|=12lambda +3mu +20 -2mu -10lambda -36`
`=2lambda +mu -16`
`Delta_(2) =|{:(1,,6,,1),(1,,10,,3),(1,,mu,,lambda):}|=10lambda +18 +mu -10 -3mu -6lambda =4lambda -2mu +8`
`Delta_(3) = |{:(1,,1,,6),(1,,2,,10),(1,,2,,mu):}|=2mu +10 +12 -12 -mu -20 =mu -10`
Thus the system has unique solutions if `Delta ne 0 " or " lambda ne 3` and the system has infinite solution if `Delta=Delta_(1) =Delta_(2) =Delta_(3) =0 " ot " lambda =3` and `mu=10`. System has no solution if `Delta =0` and at least one of `Delta_(1) ,Delta_(2) ,Delta_(3)` is nonzero of `lambda =3 " and " mu ne =10`