# How to parameterize a solution of a system of equations?

Give the parameterized solution for the dependent system represented by the matrix: $$\left(\begin{array}{ccc|r} 1 & 0 & 3 & 6\\ 0 & 1 & 2 & -4\\ 0 & 0 & 0 & 0 \end{array}\right).$$

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It is not clear what your matrix is. Is it simply the row vector $[1, 0, 3, 6, 0, 1, 2, -4, 0, 0, 0, 0]$? – Fredrik Meyer Feb 7 '11 at 5:07
Please phrase your questions like questions, not like orders. If you also state what you have tried, or why you are stuck, that usually helps. And, of course, as Fredrik says, you haven't given a matrix, you've given a sequence of numbers. – Arturo Magidin Feb 7 '11 at 5:24
also, is this homework? – InterestedGuest Feb 7 '11 at 5:36

$\begin{bmatrix} 1 & 0 & 3 & | & 6\\ 0 & 1 & 2 & | & -4\\ 0 & 0 & 0 & | & 0 \end{bmatrix}$
The way to go here is from the bottom to up. You see that the third coordinate can be anything (it's a free variable), since $x_{3}$ times $0$ equals $0$ in all cases (I am looking at the third coefficient of the third equation in your system). So let $x_{3}=t$. Then the second equation is of form $0x_{1}+1x_{2}+2t=-4$. So, $x_{2}=-4-2t$. And, finally, the first equation can be represented as $1x_{1}+0x_{2}+3t=6$, which amounts to $x_{1}=6-3t$. So your solution vector is $\begin{bmatrix} 6-3t\\ -4-2t\\ t \end{bmatrix}$.