# Linear system of equations with parameters

1. Nov 14, 2013

### MSG100

I have big problems with an equation system. It's been a long time since I worked with these type of problems and it would be wonderful if I could get an full solution of this or some similar problem.

Decide for all real a and b, the number of solutions to the equation system.

I begin with finding the determinant and get 2a-2
Is this even the right way to start?

Should I put a=1 into the equation and use Gauss elimination?

2. Nov 14, 2013

### pasmith

If the determinant (which depends only on a) is non-zero, then there is a unique solution for any b.

If the determinant is zero, then depending on b it may be that there are no solutions or infinitely many solutions.

So your first step is to calculate the determinant. The answer 2a - 2 must be wrong, because if a = 0 then the sum of the first two rows is twice the third row, and the determinant must vanish.

3. Nov 14, 2013

### MSG100

I missed to insert zero for the missing "y" in the third row.
With them in place I got it to 1a

4. Nov 14, 2013

### MSG100

I searching in my books and internet after similar problems but can't find any. I don't really understand the problem.

Here's how I evaluate my determinant. I use the Rule of Sarrus.

5. Nov 14, 2013

### Staff: Mentor

I apologize - a is the correct value. I wrote a 2 where I should have written a 1.

6. Nov 14, 2013

### MSG100

I have noticed myself that it's very easy to make small mistakes with these kind of calculations. I'm glad that I could get it right for once.

How do I go from on from here?

7. Nov 14, 2013

### Staff: Mentor

If the determinant is nonzero, the system of equations has a unique solution.

8. Nov 14, 2013

### MSG100

Hmm this is really confusing. I thought a was kinda good at Gaussian elimination but this proved me wrong:)

I put a=0 an get this:

9. Nov 14, 2013

### Staff: Mentor

The problem, as you stated it, doesn't ask for the solutions - just how many there are for various values of a. If a is 0, there are multiple solutions (actually, an infinite number of them).

10. Nov 14, 2013

### Ray Vickson

Don't put anything anywhere; just start doing Gaussian elimination. For example, the third equation gives you z = 2-x. Put that into the first two equations, to get a new 2x2 system involving x and y only. Now you can use one of these new equations to (for example) solve for y in terms of x; putting that into the other equation will leave you with an equation of the form C*x=c, where C and c involve a, etc. It is now easy to tell if that last equation has (i) a unique solution; (ii) no solutions; or (iii) infinitely many solutions.

In my opinion, getting the determinant first is a complete waste of time, because after you have obtained it you still need to start over again to tell whether you have case (ii) or case (iii). Why not just do it immediately? (Others will disagree, of course.)

11. Nov 14, 2013

### Staff: Mentor

That's a good point. You can't tell from the determinant whether there are no solutions or an infinite number of them. All you can tell is whether there is a single solution or not.