Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned}2\\1 \\0 \end{aligned}\right], \left[\begin{aligned}-3\\0\\1 \end{aligned}\right].$$
(b) The dimension of the solution space is $2$.
Work Step by Step
The coefficient matrix is given by
$$
\left[ \begin {array}{ccc} 1&-2&3\\ -3&6&-9
\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{ccc} 1&-2&3\\ 0&0&0
\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned} x -2y+3z&=0\\
\end{aligned}.
$$
The solution of the above system is $x=2s-3t$,$y=s$, $z=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form
$$x= \left[\begin{aligned} x\\ y\\z \end{aligned}\right]= \left[\begin{aligned}2s-3t\\s \\t \end{aligned}\right] =s\left[\begin{aligned}2\\1 \\0 \end{aligned}\right] + t \left[\begin{aligned}-3\\0\\1 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned}2\\1 \\0 \end{aligned}\right], \left[\begin{aligned}-3\\0\\1 \end{aligned}\right].$$
(b) The dimension of the solution space is $2$.
