# Linear Algebra System Consistency problem

1. Jan 20, 2010

### Precursor

1. The problem statement, all variables and given/known data
What relationship must exist between the constants a, b, and c for the following linear system to be consistent? What is the solution of the system when it is consistent?

$$x + y + 2z = a$$
$$2x + 2z = b$$
$$3x + y + 4z = c$$

3. The attempt at a solution
This problem wasn't something we discussed in class so I didn't really have a clear idea of going about it. So I tried to reduce the system to RREF and see what the solutions were in terms of a, b, and c, as shown below:

$$1......1......2......a$$
$$0......1......1......(a - b/2)$$
$$0......2......2......(3a - c)$$

Apparently, if I do any further reduction I will eventually get a row with all zeros on the left side, which means the system has no solutions. Obviously, this is not the correct method, because the system must have solutions for their to be any relationship between the constants. Any help on this question is greatly appreciated.

Last edited: Jan 20, 2010
2. Jan 20, 2010

### boboYO

No, remember you are not solving for x,y,z. You are finding values of a,b,c such that the system -will- have a solution.

If you get a row like 0..0..0..7a+b+3c [for example] , what is the restriction on a,b,c such that the system has a solution?

Last edited: Jan 20, 2010
3. Jan 20, 2010

### HallsofIvy

If you do further reductions you will get a last line of
$$0.....0.......0........a+ b- c$$

That will have no solution if and only if the number of the right, a+ b- c, is not 0.

4. Jan 20, 2010

### Precursor

So I have to solve for $$a+ b- c = 0$$ by using this equation with the other two?

5. Jan 21, 2010

### Precursor

Ok, so I did further reduce the matrix and did get $$a+ b- c= 0$$.
How do I continue with this to find the solution of the system?

From the reduced matrix, I was also able to obtain the following equations:
$$x+ z= b/2$$
$$y+ z= a- b/2$$

$$x= -z+ b/2$$
$$y= a-z- b/2$$

Are these the two solutions to the system?

6. Jan 21, 2010

### Precursor

I feel that I have not incorporated $$a+ b-c= 0$$ into my solution set. Am I right?

7. Jan 21, 2010

### Precursor

Any suggestions?

8. Jan 21, 2010

### vela

Staff Emeritus
Not exactly. A solution is a 3-tuple (x,y,z) that satisfies the original equations. It's possible for a system of equations to have no solution, a unique solution, or an infinite number of solutions. In this problem, you've run into the first and third cases.

If $a+b-c\ne 0$, the last line of the reduced matrix would have given you the equation

$$0x+0y+0z=\mbox{a non-zero number}$$

There are no possible values for x, y, and z that would work, so the system has no solution. For the system to have a solution then, the relationship $a+b-c=0$ must be met.

Assuming that requirement was met, you went on to find

$$x=-z+b/2$$
$$y=-z+a-b/2$$

Now suppose z=t, where t is a real number, then you'd have

$$x=-t+b/2$$
$$y=-t+a-b/2$$
$$z=t$$

For a specific value of t, you can calculate what x, y, and z are, and (x,y,z) will be a solution to the original equations. Since there is an infinite number of possible values of t, there is an infinite number of solutions to the original equations.

9. Jan 21, 2010

### Precursor

So, as you have extended it, the solution set is
$$x= -t+ b/2$$
$$y= -t+ a- b/2$$
$$z= t$$

in which an infinite amount of solutions exists. Thanks for the help.

Last edited: Jan 22, 2010
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