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!

  • $\begingroup$ Look at the system $x=0$, $x=1$, $x+y+z+w=2012$. $\endgroup$ – André Nicolas Mar 6 '12 at 23:14
  • $\begingroup$ When you talk about number of variables, make sure that the variables are independent from one another, naming alone is not sufficient. $\endgroup$ – NoChance Mar 6 '12 at 23:17
  • $\begingroup$ @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. $\endgroup$ – Chris Mar 6 '12 at 23:35
  • $\begingroup$ @EmmadKareem: Doesn't this result 'w ( 335/21, -2596/147, 104/147, 1)' clarify that the variables are not independent? :S $\endgroup$ – Chris Mar 6 '12 at 23:36
  • $\begingroup$ @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. $\endgroup$ – André Nicolas Mar 6 '12 at 23:45

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).

  • $\begingroup$ 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? $\endgroup$ – Chris Mar 6 '12 at 22:55
  • $\begingroup$ 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) ? $\endgroup$ – Chris Mar 6 '12 at 23:16
  • $\begingroup$ (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. $\endgroup$ – Gerry Myerson Mar 6 '12 at 23:43
  • $\begingroup$ (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? $\endgroup$ – Chris Mar 7 '12 at 1:20
  • $\begingroup$ 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? $\endgroup$ – Gerry Myerson Mar 7 '12 at 1:42

There isn't always a solution. You could have:

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

  • $\begingroup$ 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? $\endgroup$ – Chris Mar 6 '12 at 22:43
  • $\begingroup$ @Chris, see the answer I just posted. $\endgroup$ – Gerry Myerson Mar 6 '12 at 22:44

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