System of equation, condition of a

**arbolis** · September 14th 2008, 09:21 PM

For which values of $a$ the following system of equations has a unique solution? Infinite solutions?

$\left[ \begin{array}{ccc}x-y+z=2 \\ ax-y+z=2 \\ 2x-2y +(2-a)z=4a \end{array} \right]$ .
I've solve the system via matrices and got that $\left[ \begin{array}{ccc}x=0 \\ y=-2-4a \\ z=-4a(a-1) \end{array} \right]$ . I'm stuck from here... I see that for a given $a$ there is always a unique solution, but that you can give whatever value for $a$ so that you have an infinity of solutions. So how do I answer the question?

**Soroban** · September 14th 2008, 10:10 PM

Hello, arbolis!

For which values of $a$ does the following system of equations has a unique solution?
Infinite solutions?

$\begin{array}{ccc}x-y+z\;=\;2 & [1]\\ ax-y+z\;=\;2 & [2]\\ 2x-2y +(2-a)z\;=\;4a & [3]\end{array}$

Just "eyeballing" the system, I can see the following:

If $a = 1$ , equations [1] and [2] are identical.
. . The system will have infinite solutions.

If $a = 0$ , equation [3] becomes: . $2x - 2y + 2z \:=\:0 \quad\Rightarrow\quad x - y + z \:=\:0$
. . which is incompatible with equation [1].
The system will have no solution.

**arbolis** · September 15th 2008, 06:59 AM

Thank you Soroban!
You are right, it seems more easy than I thought. So what I did with matrices was not necessary and only complicates things.
But I'd like to know if there is a formal way to determine what values of $a$ would make the system to get only one solution. Is that possible? Or have I to "eyeball" this?

**Moo** · September 15th 2008, 12:31 PM

Hello,

Here is my trial... lol

The system can be rewritten this way :

$\begin{pmatrix} 1 & -1 & 1 \\ a & -1 & 1 \\ 2 & -2 & 2-a \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 2 \\ 2 \\ 4a \end{pmatrix}$

I don't have my notes with me, so I can't check. The wikipedia article may help you, I'm not sure (System of linear equations - Wikipedia, the free encyclopedia).

In any event, the span has a basis of linearly independent vectors that do guarantee exactly one expression; and the number of vectors in that basis (its dimension) cannot be larger than m or n, but it can be smaller. This is important because if we have m independent vectors a solution is guaranteed regardless of the right-hand side, and otherwise not guaranteed.

Independent vectors mean that there is no linear combination of two of them giving the third. Or any relation of proportionality.
If you find dependent vectors, it is likely that you'll get an infinity of solutions. And if there are dependent vectors, the discriminant will be = 0 (to be checked).

General behavior
The solution set for two equations in three variables is usually a line.
The solution set for two equations in three variables is usually a line.

In general, the behavior of a linear system is determined by the relationship between the number of equations and the number of unknowns:

1. Usually, a system with fewer equations than unknowns has infinitely many solutions.
2. Usually, a system with the same number of equations and unknowns has a single unique solution.
3. Usually, a system with more equations than unknowns has no solution.

This may be helpful too.

I'm sorry not to be able to help more, algebra is not what I prefer so I easily forget these things

Hope you'll extract something from this.

**arbolis** · September 15th 2008, 01:01 PM

And if there are dependent vectors, the discriminant will be = 0 (to be checked).

What do you mean by discriminant? Anyway, I've not yet seen basis (even if I had to deal with them when I studied Numerical Analysis).
And I knew about

1. Usually, a system with fewer equations than unknowns has infinitely many solutions.
2. Usually, a system with the same number of equations and unknowns has a single unique solution.

.
And about

The system can be rewritten this way :

I know, that's what I did to reach

but I certainly did an error since it doesn't fit with the eyeball of Soroban.
So my question remains "I'd like to know if there is a formal way to determine what values of

would make the system to get only one solution."
Thanks for your time Moo and your help, I appreciate the research you've done for me.
Lastly,

I'm sorry not to be able to help more, algebra is not what I prefer so I easily forget these things

I hate this matter (even if less than a year ago) but it is an extremely important one. Not only for me (as a physics student) but for many other careers. I had an exam of physics I last Wednesday and I totally depreciated Algebra II. The exam of Algebra II is coming this Wednesday so I'm hurrying up!

**Moo** · September 15th 2008, 01:11 PM

Originally Posted by arbolis

What do you mean by discriminant?

Determinant

Sorry, I'm being very tired..

I hate this matter (even if less than a year ago) but it is an extremely important one. Not only for me (as a physics student) but for many other careers. I had an exam of physics I last Wednesday and I totally depreciated Algebra II. The exam of Algebra II is coming this Wednesday so I'm hurrying up!

So good luck

There is (at least) a little mistake in your answer.

You should have had (1-a)x=0. But this doesn't mean that x=0 ! This is the first eyeball

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