# Matrices and Systems of Linear Equations

1. Mar 9, 2016

### DiamondV

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

2. Relevant equations

3. The attempt at a solution

No clue really.

I went ahead and tried to simplify this by turnining it into an echelon matrix.
But Im sort of stuck now, since I cant divide by -k because I don't know whether or not it is equal to 0?

Last edited: Mar 9, 2016
2. Mar 9, 2016

### Staff: Mentor

Your last operation, replacing row3 by row3 + row2, was really a wasted effort, since it didn't eliminate the leading entry of row 3.

Look at the preceding matrix, especially the similarity of row2 and row3. What must happen so that the matrix represents a system with multiple solutions?

3. Mar 9, 2016

### DiamondV

You would need to assign a parametric value to one of the variables in order to get more than one solution.
The preceeding matrix has an equation from row 2: -y+z=2 and from row 3: (1-k)y +z = 2

4. Mar 9, 2016

### epenguin

Oh dear, at school I would have counted that as a pretty easy set of simultaneous equations to solve. If introducing matrices makes something like this difficult for students, I'm inclined to wonder if they are better off without them?!

In your position my inclination would hbave been to solve it the easy way, and then reformulate it in the (equivalent) matricial terms they want, and pretend I did it that way! It might even give a better idea of what the matrices are all about and how they work. But you may have been trained up a different way.

Last edited: Mar 9, 2016
5. Mar 9, 2016

### Staff: Mentor

I think you have this the wrong way around.

If you have more than one solution, this can happen in one of three ways:
All three planes intersect in a line.
Two of the planes intersect in a line, and the third plane is identical to one of the other two.
All three planes are coplanar (they're all the same plane).

Focus on the geometry of this situtation, and not the matrix. The above should be a hint on how to deal with what you have below.

6. Mar 9, 2016

### Ray Vickson

You echo my sentiments exactly. I am constantly nagging in this Forum about how much easier it is to do such problems directly, as though matrices had never been invented.

7. Mar 9, 2016

### SteamKing

Staff Emeritus
Or, you can learn which conditions in the system of equations give rise to the different number of solutions.

http://www.mathwarehouse.com/algebra/linear_equation/systems-of-equation/index.php

The question asks what value of k gives more that one solution to the given system of equations. What the OP apparently doesn't realize is that a system of linear equations can have 0 solutions, exactly 1 solution, or infinitely many solutions, and that each of these conditions can be determined by examining the matrix of coefficients. It also helps to calculate the determinant of the matrix of coefficients.

It is not necessary to solve the system in order to figure out how many solutions there are.

8. Mar 10, 2016

### Ray Vickson

No, but in the case of 0 or (infinitely) many solutions, the determinant of the matrix is not good enough; the right-hand-side also comes into play. Again, while I, personally, WOULD use matrices for such problems, it is only because I know the subject already and so do not need to study/learn it. I am still convinced that using matrices at the beginning is not the best way---reserve them for later, when they really matter.

Last edited: Mar 10, 2016
9. Mar 12, 2016

### DiamondV

To be fairly honest, these seem so pointless to me, all our lecturer says to us is that they are very important in advanced engineering and we need to know it really well. So I will probably need them later on but right now, its just another tool im adding to my bank of tools that are supposedly very important later on.

10. Mar 12, 2016

### ehild

You can rearrange the system of equations, making it easier to turn into an echelon matrix.
x + 2z + y = 3
z - y = 2
x + z + ky = 1

Try, what do you get?

11. Mar 13, 2016

### epenguin

'Will be useful later on' is not very efficient as a motivator. But but it is very often met. The other departments say to the maths department, you teach this, we can't teach the physics or whatever without it, or we don't want to have to stop and explain. And maybe there are ways to integrate them, you might think maths and science could be taught in some single integrated course, but doing that is not as easy as saying it.

While when the 'later on' comes round later on, the student has likely forgotten the useful thing he unmotivatedly stumbled through. But hopefully something has sunk in, and can be dug up and rendered useful. What can you do? Just treated as a challenge for the moment I guess.

? It seemed to me that you had been thrown by the 'more than one solution'. Although this was clearly an exercise on something you had had in your course or book - but you were not recognising this. For linear systems like the kinds in your exercises the solutions are either none (inconsistent equations, usually something wonky in the formulation from any real world problem) or one, or an infinite number. But instead of saying an infinite number, I've noticed the mathematicians often seem to say 'more than one' and this throws the students because although infinity is certainly more than one the phrase suggests 'several', say two, three or four like polynomial equations the student has met. So the expression is true but misleading. I'm just wondering and guessing that the reason they say this is that the mathematicians are into far more general things with linear spaces, and perhaps for all I know in some of their fancy spaces it is a finite number of solutions. But in that case it would IMHO be better that the few unlearn later than that the many are obstacled for learning earlier.

12. Mar 13, 2016

### CrazyNinja

Check your last step once. Get that right and the answer will present itself.