# Finding The Constants a,b and c.

1. Jul 30, 2013

### Bashyboy

1. The problem statement, all variables and given/known data
In Exercises 49 and 50, find values of a, b, and c(if possible) such that the system of linear equations has (a) a unique solution, (b) no solution, and (c) an infinite number of solutions.

49.

$x+y ~ = 2$

$~y+z =2$

$x+ ~ z =2$

$ax + by + cz=0$

2. Relevant equations

3. The attempt at a solution

I immediately recognized this system as one consisting of planes. Hence, wouldn't it be impossible for there to exist a unique solution, as my geometric intuition of the situation ?

2. Jul 30, 2013

### hilbert2

What about case a=1, b=-1, c=0 ? Isn't there a unique solution?

3. Jul 30, 2013

### HallsofIvy

Staff Emeritus
No, it is quite possible for four planes to intersect in a single point. What makes you think otherwise?

(Boy, I have to learn to type faster!)

4. Jul 30, 2013

### Bashyboy

Well, I was considering the case of two planes. Would you agree that it is impossible for there to exist a unique solution, when considering two planes?

5. Jul 30, 2013

### HallsofIvy

Staff Emeritus
No, that's not enough. You need at least three planes. And 4 is larger than 3!

What you can do is solve the first three equations for the unique point of intersection, then see what a, b, and c must be in order that those values of x, y, and z satisfy the final equation.

6. Jul 30, 2013

### Bashyboy

HallsofIvy, I am confused, what do you mean by, "No, that's not enough. You need at least three planes. And 4 is larger than 3!" what are you alluding to?

7. Jul 31, 2013

### haruspex

That's right - with just two planes you must have no solutions, or a line of solutions, or a plane of solutions.
Considering just the first three planes you're given, how many solutions are there?

8. Aug 6, 2013

### Bashyboy

Haruspex, I took the first three equations and came up with this:

(1) $x+y=2$

(2) $y+z=2$

(3) $x+z=2$

Rearranging (1), $y=2-x$, and substituting the result into (2): $(2-x) +z=2$, which becomes $z-x=0$; adding this equation to (3): $2z=0 \implies z=0$

If z=0, then y=2 and x = 0.

If the fourth equation was equal to two, rather than 0, then I could choose the constants such that it would be identical to one of the other planes.

9. Aug 6, 2013

### Bashyboy

Hold on, I think I figured it out. Let me type my response.

10. Aug 6, 2013

### Bashyboy

For the first three exists, there exists a unique solution, namely, (0,2,0). We need to retain this uniqueness, when dealing with the fourth equation, and to do so, (0,2,0) must satisfy the fourth equation, and we must choose constants that are conducive to this. a(0) + b(-2) + c(0) = 0. From this, it is clear that, a and c can be any real numbers, but b must be zero. Does this sound correct?

11. Aug 6, 2013

### D H

Staff Emeritus
Does this satisfy x+z=2? You might want to try again.

12. Aug 6, 2013

### haruspex

Retry those steps.

13. Aug 7, 2013

### Bashyboy

Oh, wait. There should have been a 2 on the right-hand side of the equation, meaning that z=1, y=1, and x=1.

14. Aug 7, 2013

### D H

Staff Emeritus
Much better.

15. Aug 7, 2013

### Bashyboy

For part (b), to make the system inconsistent, must I chose constants that will guarantee the equation a + b + c = 0 is false? I'm not sure how I will solve part (c), yet.

Last edited: Aug 7, 2013
16. Aug 7, 2013

### HallsofIvy

Staff Emeritus
Yes, start from the first three equations: Subtracting y+ z= 2 from x+ y= 2, the "y" terms cancel and we have x- z= 0. Adding that to x+ z= 2, the "z" terms cancel and we have 2x= 2 so that x= 1. Then x+ z= 1+ z= 2 so that z= 1 and x+y= 1+ y= 2 so y= 1. The solution to the first three equations is x= y= z= 1.

NOW look at "ax+ by+ cz= 1". With x= y= z= 1, that gives a+ b+ c= 1. As long as that is true, the four equations will have the unique solution x= y= z= 1. If it is not true, there will be no solution. Since x= y= z= 1 is the only solution to the first three equations, there is no way to have more than one solution.