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I've been asked to find the general solution of this matrix:

$\left[\begin{array}{cccc} 1 & -3 & 0 & -5 \\ -3 & 7 & 0 & 9 \\ \end{array}\right]$

When you try to find the general solution, you first put the matrix into its reduced echelon form, correct?

Thus I obtained:

$\left[\begin{array}{cccc} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ \end{array}\right]$

So I've solved for $x_{1}, x_{2}$ but I don't know how to describe $x_{3}$. Is it a free variable? Since for any $x_{3} \in \mathbb{R}$ the system has a solution, correct? I'm just not sure because it's an even problem in my textbook.

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    $\begingroup$ Right, $x_3$ is free. If this is the augmented coefficient matrix then the solutions have the form $x_1=4, x_2=3$ and $x_3$ is whatever you like. $\endgroup$ – James S. Cook Jan 24 '14 at 16:56
  • $\begingroup$ $x_3$ is like the pointless button: WARNING: Pointless! $\endgroup$ – AlexR Jan 24 '14 at 16:56
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If you want to test it, simply compute $$\begin{bmatrix}1 & -3 & 0\\-3 & 7 & 0\end{bmatrix}\begin{bmatrix}4\\3\\x_3\end{bmatrix}.$$

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  • $\begingroup$ Interesting, I will probably learn this soon! Thanks. $\endgroup$ – Bobby Lee Jan 24 '14 at 17:01
  • $\begingroup$ Have you not done matrix multiplication before? $\endgroup$ – Cameron Buie Jan 24 '14 at 17:03
  • $\begingroup$ I know how to do it but I haven't seen this test before. $\endgroup$ – Bobby Lee Jan 24 '14 at 17:09
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    $\begingroup$ Well, just bear in mind that for an $m\times n$ matrix $A,$ an $m\times 1$ vector $b,$ the numbers $x_1,...,x_n$ that you derive by putting $[A\mid b]$ in reduced echelon form are precisely the entries of an $n\times 1$ vector $x$ such that $Ax=b,$ so that's always a way we can test such solutions. $\endgroup$ – Cameron Buie Jan 24 '14 at 17:13

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