Question

Show that the system of equations `3x-y + 4z = 3, x + 2y-3z =-2` and `6x + 5y + lambdaz=-3` has at least one solution for any real number `lambda.` Find the set of solutions of `lambda =-5`

Answer 1

Correct Answer - `x = (4-5k)/(7), y = (13k-9)/(7), z =k`
The given system of equations
`3x-y +4z =3`
x +2y-3z = -2
`6x+5y +lambdaz =-3`
has atleast one solution, if `Delta ne 0`.
`therefore Delta = |{:(3, -1, 4), (1, 2, -3), (6, 5, lambda):}| ne 0`
`rArr 3(2lambda + 15) +1(lambda + 18) + 4 (5-12) ne 0`
`rArr 7(lambda + 5) ne 0`
`rArr lambda ne -5`
Let z= -k, then equations become
3x-y =3-4k
and x +2y -3k =-2
On solving, we get
`x = (4-5k)/(7), y =(13k-9)/(7), z =h`