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I'm in the process of solving a system of three linear equations in three unknowns, the last row of the matrix contains all zeros. How does this affect the solution of the system of equations?

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  • $\begingroup$ Do you mean how to solve? It doesn't affect things at all; just continue solving. We may be able to get qualitative information about solutions; information about the rows tells us something about the row space.... Or just finish solving and see the behavior for yourself. $\endgroup$
    – user14972
    Commented Feb 21, 2013 at 1:44

2 Answers 2

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It means that there is a "free variable", and infinitely many solutions to the system, because that "free variable" can be assigned any value. The other variables can often be expressed in terms of the free unknown variable; that is, their values will often depend on the the unknown value, depending on how "reduced" your matrix is.

E.g. If we have the augmented coefficient matrix reduced to the following, representing a system of three equations in the three unknowns $x, y, z$:

$$ \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} $$

Then $z$ is the free variable. $y = 2$, and $x = 1 - y - z = 1 - 2 - z = -(1 + z)$. So if we let $z = \alpha$, we'd have the an infinite "family" of solutions which can be represented as $$ \begin{pmatrix}\\ \\ -(1 + \alpha) \\ \\ 2 \\ \\ \alpha \\ \\ \end{pmatrix}$$ which depends on the value of $\alpha$, which can be any of infinitely many values.

Hence, there are an infinite number of solutions.

But this still allows us to impose some constraints on the solutions: specifically, $y$ must be $2$, and whatever the value of $\alpha$, $x = -(1+ \alpha)$.

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that means the system has infinitely many solutions, and if your last matrix is in ref or rref then you will have somthing like this$$ x_1+2x_2+x_3=3\\ x_2-x_3=4 $$which implies $$x_1=3-2x_2-x_3\\ \text{and}\hspace 2mm x_2=4+x_3$$ let$x_3=t$ then you will get the value of $x_1,x_2$ and hence your solution.

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  • $\begingroup$ If $x_3=t$, then $x_2=4+t$. There is only one parameter needed here. $\endgroup$
    – Julien
    Commented Feb 21, 2013 at 2:40
  • $\begingroup$ @julien Yes you are correct. Just edited it, Thank you. $\endgroup$
    – i.a.m
    Commented Feb 21, 2013 at 2:53

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