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Let $A = \begin{bmatrix}1&-5&-3&2\\4&-20&-12&8\end{bmatrix}$

Describe all solutions of $Ax = 0$

$x = x_2 \begin{bmatrix}\\\\\end{bmatrix} + x_3 \begin{bmatrix}\\\\\end{bmatrix} + x_4 \begin{bmatrix}\\\\\end{bmatrix}$

I put the matrix $A$ in RREF form to get:

$\begin{bmatrix}1&-5&-3&2\\0&0&0&0\end{bmatrix}$

Then I saw that $x_1 = -5x_2 + 3x_3 - 2x_4 \implies x_1 = -5r + 3s - 2t$ (because $x_1$ is not a free variable right?)

Now I don't understand how to get the rest of my rows though?

I know -5 3 -2 will be the first row of each of my vectors respectively in order

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    $\begingroup$ $x_2=r$, $x_3=s$, $x_4=t$. So $$\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4} = ?$$ $\endgroup$ – user137731 Jun 24 '16 at 21:49
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    $\begingroup$ Like I commented in your last question, you should read this answer if you're still having trouble with this type of problem. $\endgroup$ – user137731 Jun 24 '16 at 21:50
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You got the reduced matrix $$ \begin{pmatrix} 1&-5&-3&2 \end{pmatrix} $$ thus one equation $$ x_1 - 5 x_2 - 3 x_3 + 2 x_4 = 0 $$ and four unknowns. We pick the last three variables as free variables and get the general solution $$ \begin{align} L &= \left\{ (5 x_2 + 3 x_3 - 2 x_4, x_2, x_3, x_4)^T \mid x_2, x_3, x_4 \in \mathbb{R} \right\} \\ &= \left\{ x_2(5 e_1 + e_2) + x_3(3 e_1 + e_3) + x_4(-2 e_1 + e_4) \mid x_2, x_3, x_4 \in \mathbb{R} \right\} \\ &= \left\{ x_2 (5, 1, 0, 0)^T + x_3 (3, 0, 1,0)^T + x_4 (-2, 0, 0, 1)^T \mid x_2, x_3, x_4 \in \mathbb{R} \right\} \\ \end{align} $$ where the $e_i$ are the standard basis vectors.

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It's trivial, figured it out thanks to @Bye_World

$x = r \begin{bmatrix}5\\1\\0\\0\end{bmatrix} + s \begin{bmatrix}3\\0\\1\\0\end{bmatrix} + t \begin{bmatrix}-2\\0\\0\\1\end{bmatrix}$

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