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We have a system of 3 linear equations with 4 variables, does it always have a solution? (homework)

In example we have this matrix (after the row reduction process):

1   0   0   -335/21 0   1   0   2596/147 0   0   1   -104/147 

All of the 3 linear equations are equal to 0.

The 1st equation is this: $x+ 0y + 0z + (-335/21)w = 0$ So the solution to this system is:

w ( 335/21, -2596/147, 104/147, 1) 

So the system, has infinite solutions?

But what is the answer to the question (in title)?
(We have to say if it's true/ false, and why)
I believe it's false as in general a system of linear equations can have infinite solutions, or 1 unique or none..is that correct?

Thank you for your time!

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    Look at the system $x=0$, $x=1$, $x+y+z+w=2012$.2012-03-06
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    When you talk about number of variables, make sure that the variables are independent from one another, naming alone is not sufficient.2012-03-06
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    @AndréNicolas: It has 3 equations and 4 unknowns and it has **no** solutions, right? So the answer must be, there is always a solution, only if the system is homogeneous.2012-03-06
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    @EmmadKareem: Doesn't this result 'w ( 335/21, -2596/147, 104/147, 1)' clarify that the variables are not independent? :S2012-03-06
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    @Chris: The answer is that sometimes there is no solution. One might volunteer that when the system is homogeneous, there is always a solution, but that wasn't the question. **Any** example, like the one that I gave, or one that you might come up with, shows that it ain't necessarily so.2012-03-06
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    @AndréNicolas: Thank you!!2012-03-07

2 Answers 2

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A homogeneous system of 3 linear equations in 4 unknowns always has a solution, in fact, always has a non-trivial solution, a solution where the unknowns are not all zero. A system is homogeneous if the constant terms are all zero, which is the situation you are describing in your question when you say "all of the three linear equations are equal to zero."

More generally: A homogeneous system of linear equations always has at least one solution, namely, the solution in which each unknown is zero. If the number of unknowns exceeds the number of equations then a homogeneous system is guaranteed to have infinitely many solutions (and thus solutions in which the unknowns are not all zero).

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    homogeneous!It was the word I was looking for (It's a greek word and I am from greece!). So the answer to the title question would be wrong? The system must be a homogeneous, to always have a solution. Right?2012-03-06
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    Also you've said that (in the case of a homogeneous system of linear equations) in a matrix of m rows and n columns, if ( m < n ) then we have infinite solutions. What if (m > n) ?2012-03-06
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    (1) right, if a system is not homogeneous, it might have a solution, or it might not. (2) if a homogeneous system has more equations than unknowns it is still guaranteed to have the solution where all the unknowns are zero; it might, or might not, have other solutions. You should try to make up some examples to illustrate the possibilities.2012-03-06
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    (2)We have a question "what if we were to add 2 equations at the system of the 3 linear equations (and now we have 5 equations (rows) and 4 unknowns (columns), we would have exactly one solution to the system". So as you said, we have the solution where all the unknowns are zero, but what steps should I follow to find out if there are more solutions?2012-03-07
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    There is a standard procedure for solving systems of linear equations (homogeneous or not), sometimes going by the name of Gaussian elimination, involving the performance of row operations on the matrix of coefficients. Have you not learned this procedure?2012-03-07
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    I know the Gaussian elimination process, I didn't make myself clear. What I don't know is which 2 equations to add? How should I "select" these 2 equations?2012-03-07
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    I would expect that adding the equations $x=0$ and $y=0$ would result in a situation where the only solution is the one with all the variables zero.2012-03-07
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    What if we added x = c, and then w would be = a*c and y = b*w, z = t*w (all the other variables will have be constants). Now if after this we add the 2nd equation (i.e. y = c + 1) the system will have no solution...right?2012-03-07
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There isn't always a solution. You could have:

$$ x+y+w+z=1 $$ $$ x+y+w+z=2 $$ $$ x+y+w+z=3 $$

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    So the question is general and is not confined to the systems of linear equations that are equal to 0, right? If it was confined to the systems of linear equations that are equal to 0, what would be the answer?2012-03-06
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    @Chris, see the answer I just posted.2012-03-06