Answer
In parametric vector form, the solution is
$=x_2\left[\begin{array}{ c }2\\1\\0\\0\\0\\0\end{array}\right]+x_3\left[\begin{array}{ r }-3\\0\\1\\0\\0\\0\end{array}\right]+x_5\left[\begin{array}{ r }-29\\0\\0\\-4\\1\\0\end{array}\right]$
Work Step by Step
$\left[\begin{array}{ r r r r r r r }1&-2&3&-6&5&0&0\\0&0&0&1&4&-6&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&0\end{array}\right]\sim \left[\begin{array}{ r r r r r r r }1&-2&3&-6&5&0&0\\0&0&0&1&4&0&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&0\end{array}\right]\sim \left[\begin{array}{ l c l l l l l }(1)&-2&3&0&29&0&0\\0&0&0&(1)&4&0&0\\0&0&0&0&0&(1)&0\\0&0&0&0&0&0&0\end{array}\right]$
$\begin{array}{l}\left(x_1\right)-2x_2+3x_3+29x_5=0\\(x_4)+4x_5=0\\(x_6)=0\\0=0\end{array}$
The basic variables are$ x_1, x_4,$ and$ x_6$; the free variables are$x_2, x_3$, and$ x_5$. The general solution is $x_1=2x_2-3x_3-29x_5,x_4=-4x_5,\text{ and }x_6=0$. In parametric vector form, the solution is
$x=\left[\begin{array}{ c }x_1\\x_2\\x_3\\x_4\\x_3\\x_6\end{array}\right]=\left[\begin{array}{ c }2x_2-3x_3-29x_5\\x_2\\x_3\\-4x_5\\x_5\\0\end{array}\right]=\left[\begin{array}{ c }2x_2\\x_2\\0\\0\\0\\0\end{array}\right]+\left[\begin{array}{ c }-3x_3\\0\\x_3\\0\\0\\0\end{array}\right]+\left[\begin{array}{ c }-29x_5\\0\\0\\-4x_5\\x_5\\0\end{array}\right]$
$=x_2\left[\begin{array}{ c }2\\1\\0\\0\\0\\0\end{array}\right]+x_3\left[\begin{array}{ r }-3\\0\\1\\0\\0\\0\end{array}\right]+x_5\left[\begin{array}{ r }-29\\0\\0\\-4\\1\\0\end{array}\right]$
