Define coefficients for system of equations with 3 variables

1. Sep 9, 2009

thecokeguy

1. The problem statement, all variables and given/known data
I have the following system og equations:

$\left\{\begin{array}{l} y + 2z = 1 \\ x + 4y + 3z = C \\ x + y + Bz = B \end{array}\right.$

a) Create the augmented matrix and reduce to row echelon form (I've already done this)
b) For which values of A and B is the system inconsistent?
c) For which values of A and B does the system have only one solution?
d) Determine the complete solution to the system for all values of A and B.

2. Relevant equations
The system in row echelon form:
$\begin{pmatrix} 1 & 4 & 3 & C \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & \frac{B - C + 3}{B + 3} \end{pmatrix}$

3. The attempt at a solution
Earlier when solving this kind of problems with systems with 2 variables, I solved it geometrically as lines being parallel, intersecting etc. Of course this could be solved geometrically too as planes, but as I can't imagine how to do it geometrically with >3 variables, I thought I've missed something.

If someone could point me in a direction, then I'll figure out the rest myself.

Thanks.

2. Sep 9, 2009

njama

Ok, now go back of the system of equations. You got:

$$\left\{\begin{matrix} x+4y+3z=C\\ y+2z=1\\ z=\frac{B-C+3}{B+3} \end{matrix}\right.$$

What if B=-3 ?

3. Sep 9, 2009

thecokeguy

As that's not possible, I'll guess the system would be inconsistent?

4. Sep 9, 2009

njama

Yup, you're right.

Now find x,y,z solutions and again tell me for which is inconsistent and for which values its consistent.

5. Sep 9, 2009

thecokeguy

I'm not sure what you want... Do you want me to back substitute to get expressions for y and x too, and then define values for B and C, which results in division by 0?

6. Sep 9, 2009

njama

Yes, you need to do back substitution.

But where is A ? I can't see any A in the problem.

7. Sep 9, 2009

thecokeguy

There isn't any A... Only B and C... Don't know why

8. Sep 9, 2009

njama

Maybe C=A ? Anyway, do the back substitution and find the values of x,y,z in terms of B,C.

9. Sep 11, 2009

thecokeguy

I don't see the relevance as x and y would only incorporate multiples of z, which would be the same fraction and still be inconsistent for B = -3 and thereby be consistent for B not equal to -3?

10. Sep 12, 2009

njama

Yes, that's completely correct.

11. Sep 13, 2009

thecokeguy

After 5 hours of searching. I finally found an example I could understand a general solution in.

$z=\frac{B-C+3}{B+3} \iff z(B+3) - B + C - 3 = 0$

With the equation above, it's easy to pick the values for B and C.

b) No-solutions: B=-3 and C!=0
c) One-solution: B!=-3
d) Complete solution: B=-3 and C=0 ... (-4+5z, 1-2z, z)

Is this completely wrong?

12. Sep 13, 2009

njama

That's correct again.

For d) I would rather say infinite many solutions.

13. Sep 13, 2009

thecokeguy

Just using the problems own words... As it's written by my professor and I like high grades, that's what I useually do :P

But thanks a lot... The feedback was just what I needed and much appreciated.

14. Sep 13, 2009

njama

You're welcome!