System of Equations - Question Help (Uni level)

Hi there would really appreciate any help on this question. I have no idea how to do it...i have a feeling you need to reduce the equations using gaussian method or guass jordan (not quite sure what the difference is between the two) but any suggestions would be so helpful!! Panicking because my exam is so soon.

Question

Consider the following set of linear equations:

- x + y + 2z = 2

x + 0y + az = -1

ax + 3y + 10z = 7

Find for which value of the parameter "a" will the system have:

1) exactly one solution
2) many solutions
3) no solutions

B) Find the solution space for the case where there are many solutions.

Thanks so much! any help would be really appreciated :smile:

Reply 1

DFranklin

Despite what you've said about "uni level", in practice it's hard to do better here than to solve this by "A-level" methods.

That is, try to solve the simultaneous equations, there will be times when you want to divide; when you do so, you need to think about whether what you are dividing by could be 0, and if so, whether this leads to no solutions (because you end up with something like 0x = 1), or infinite solutions (because you end up with something like 0x = 0).

Reply 2

FireGarden

If you rewrite your equations as a single matrix equation, take the coefficient matrix and find the values of a that make the matrix non-singular. These values of a are the ones that give only one solution.

Likewise, values of a that make the coefficient matrix 0 mean the system has either no or infinitely many solutions.. but I'm not sure how to tell them apart.

Reply 3

div curl F = 0

Original post by FireGarden

Likewise, values of a that make the coefficient matrix 0 mean the system has either no or infinitely many solutions.. but I'm not sure how to tell them apart.

Under these circumstances you have to calculate the "values" of x, y and z. For instance, if you put your value of "a" into the system of equations, you'll only end up with two distinct equations at most (the degeneracy in the equations is the reason why the determinant is zero). An example of these equations would be

x + z = 3, \;\; y - z = 2.

Here we can choose any value we please for z, say:

z = k \in \mathbb{R}

. This then fixes the allowed values of x and y, i.e.

x = 3 - k , \;\; y = 2 + k

, and leads to infinitely many valid solutions.

Reply 4

DFranklin

Original post by FireGarden

For those cases, you essentially need to do what I suggested (or at least, that is how I was taught).

In other words if you only need to tell whether or not there's a unique solution, by all means use matrix or determinant methods.

But if (in the singular case) you need to tell whether there are 0 or infinite solutions, you're going to end up solving the equations by hand (for that case) anyhow, so you might as well just do that in the first place.

Reply 5

guy_231

thanks guys! yeah it definitely has to be turned into a matrix and then reduced. ill give it a go!

Reply 6

DFranklin

Original post by guy_231

thanks guys! yeah it definitely has to be turned into a matrix and then reduced. ill give it a go!

I do love it when people ignore every word I say because it doesn't tally with what they want to hear.

FWIW, the method I suggest is the one recommended to me by the Cambridge Tripos Chief Examiner, back in the day.