Hey guys,
I have to solve the following problem:
Let the following systems of equations be given:
i)
2x 1 − 3x 2 + x 3 = 4
x 1 − 4x 2 − x 3 = 2
− 3x 1 + tx 2 − 2x 3 = − 3
ii)
x 2 − x 3 = 2
− x 1 + 2x 2 − 4x 3 = 3
x 1 − x 2 + 3x 3 = t + 1
For which values of t does the system have
a) a unique solution? Find this solution.
b) inﬁnitely many solutions? Find the solution set.
c) no solution?
So I started with the first one and get
x2 = -7/(8+t)
so I guess c) if t=8 the system has no solution.
For the others I just don't know how to proceed... Thanks for any help. :)
Thank you ibdutt. This is very helpful but using this method in this special case I get:
for no solution: 5/(8+t) = 3/0
for infinitely many solutions: 5/(8+t) = 3/0 = 0/7
for unique solution: 5/(8+t) /= 3/0
....??
which is the same as 0(5)= 2(8+ t). What does that tell you?
Which is the same as 3(7)= 0. What does that tell you?for infinitely many solutions: 5/(8+t) = 3/0 = 0/7
Which is the same as 3(8+t) is NOT equal to 0. What does that tell you?for unique solution: 5/(8+t) /= 3/0
....??
well maybe that there is no solution for t = -8 and a unique solution for t /= -8 and that there are no infinitely many solutions..?
But if all t /= -8 have a unique solution I would think that there are infinitely many?!