# Convert Equations To Aug Matrix And Solve

1. Jan 23, 2010

### maherelharake

1. The problem statement, all variables and given/known data

For what value(s) of 'k' will the system have (a) no solution, (b) a unique solution, and (c) infinitely many solutions?

x-2y+3z=2
x+y+z=k
2x-y+4z=k^2

2. Relevant equations

3. The attempt at a solution

I worked at the matrix and this is what I have so far. I have no idea if I made a mistake along the way, and if not, I am curious on if I thought about the actual answers correctly. I will attach my work so far. Thanks in advance. I will respond to any advice quickly.

#### Attached Files:

• ###### Scan.pdf
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25.5 KB
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2. Jan 23, 2010

### rasmhop

I haven't confirmed your calculations, but I assume that they are correct (except for your guesses for a,b,c). Using row operations on the matrix won't change the solution set so try subtracting the third row an appropriate number of times from row 1 and 2 such that your matrix is of the form:
$$\left[\begin{array}{ccc|c} 1 & 0 & 0 & f(k) \\ 0 & 1 & 0 & g(k) \\ 0 &0 & 1 & h(k) \end{array} \right]$$
For some functions f,g,h. Then you get the unique solution: x = f(k), y = g(k), z = h(k).

By the way, for this reason it would have been sufficient to row reduce the matrix of coefficients and see if you could get it in the form $I_3$ (3x3 identity matrix).

3. Jan 23, 2010

### maherelharake

Thanks for responding. I was not able to make progress to get the form that you state above. What I attached was as far as I could get, unfortunately.

4. Jan 23, 2010

### Staff: Mentor

You didn't show your intermediate work, but the matrix you ended up with is incorrect. In my second step I got this augmented matrix:
Code (Text):

[1 -2 3 | 2]
[0 3 -2 | k-2]
[0 3 -2 | k^2-4]

The next step would be to eliminate the first three entries in the third row, producing a row of zeroes in its first three entries.

It looks to me that for some values of k there will be no solutions, and for other values of k, there will be an infinite number of solutions. Also, for no value of k is there a unique solution.

5. Jan 23, 2010

### maherelharake

Aw man I see where I made a mistake. It was on the number in the third column and the third row. Let me try and redo my calculations and I will upload my progress ASAP. Thanks.

6. Jan 23, 2010

### maherelharake

I have reuploaded my updated progress. Wouldn't the solution hold if k=-2 or k=1?

#### Attached Files:

• ###### Scan.pdf
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7. Jan 23, 2010

### Staff: Mentor

You have a sign error last row of your final matrix, so these aren't the right values of k. Also, what do you mean by saying "Wouldn't the solution hold..."? What solution are you talking about? The problem is to find values of k for which the system has no solution, a unique solution, and an infinite number of solutions. That's all the problem is asking for.

8. Jan 23, 2010

### maherelharake

Oh ok so it's supposed to be (k-2)(k-1) in the bottom row. Sorry about the wording, that was a bad way for me to put it. Would it be an infinite number of solutions for k=-1? Also would it be no solution for k=2? And I'm not sure about a unique solution. Thanks again for all your help, and sorry I took so long to respond to your last response.

9. Jan 24, 2010

### Staff: Mentor

No it's not. One of those factors is correct, but the other one isn't.

Code (Text):

[0 0 0 | <something>]

The system of equations represented by the augmented matrix is inconsistent if <something> is not equal to zero. In that case, what you have is 0x + 0y + 0z = <nonzero value>, which is impossible.

On the other hand, the system is consistent if <something> = 0, but since you really have only two equations in three unknowns, the system is underdetermined and you have an infinite number of solutions.

These are the only two possibilities, so it's not possible for this system of equations to have just a single, unique solution.

Think about what values (plural) of k make <something> equal to zero. For those values of k, there will be an infinite number of solutions to the system of equations.

For any other values of k, <something> is nonzero, and the system is inconsistent; hence the system has no solutions.

10. Jan 24, 2010

### maherelharake

Ahhh dumb mistake. (k-2)(k+1) is what it's supposed to be.
So I am assuming that if k=2 or k=-1, the system has infinite solutions and if k≠2 or
k≠-1 the system has no solutions?

11. Jan 24, 2010

### Staff: Mentor

Almost perfect. If x ≠ 2 AND x ≠ -1, the system has no solutions.

12. Jan 24, 2010

### maherelharake

Ohhhh right. Got it! Thanks a lot sir. I really appreciate it.

13. Jan 24, 2010

### Staff: Mentor

You're welcome. Hang in there!