Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned} -\frac{3}{8}\\ \frac{9}{8} \\ \frac{1}{8}\\1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $1$.
Work Step by Step
The coefficient matrix is given by
$$
\left[ \begin {array}{cccc} 2&2&4&-2\\ 1&2&1&-2
\\ -1&1&4&-2\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{cccc} 1&0&0&\frac{3}{8}\\ 0&1&0&-{
\frac {9}{8}}\\ 0&0&1&-\frac{1}{8}\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned}
x_1 + \frac{3}{8}x_4&=0\\
x_2-\frac{9}{8}x_4&=0\\
x_3-\frac{1}{8}x_4&=0
\end{aligned}.
$$
The solution of the above system is $x_1=-\frac{3}{8}t$,$x_2=\frac{9}{8}t$,$x_3=\frac{1}{8}t$, $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form
$$x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}-\frac{3}{8}t\\\frac{9}{8}t \\\frac{1}{8}\\t \end{aligned}\right] = t\left[\begin{aligned} -\frac{3}{8}\\ \frac{9}{8} \\ \frac{1}{8}\\1 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned} -\frac{3}{8}\\ \frac{9}{8} \\ \frac{1}{8}\\1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $1$.
