Define coefficients for system of equations with 3 variables

[thecokeguy]

Homework Statement

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.

Homework Equations

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

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.

 

[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 ?

 

[thecokeguy]

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

 

[njama]

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

Yup, you're right.

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

 

[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?

 

[njama]

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?

Yes, you need to do back substitution.

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

 

[thecokeguy]

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

 

[njama]

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

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

 

[thecokeguy]

Yup, you're right.

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

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?

 

[njama]

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?

Yes, that's completely correct.

 

[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?

 

[njama]

That's correct again.

For d) I would rather say infinite many solutions.

 

[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.

 

[njama]

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.

You're welcome!