Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right], \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $2$.
Work Step by Step
The coefficient matrix is given by
$$
\left[ \begin {array}{cccc} 1&-3&1&1\\ 2&1&-1&2
\\ 1&4&-2&1\\ 5&-8&2&5\end {array}
\right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{cccc} 1&0&-\frac{2}{7}&1\\ 0&1&-\frac{3}{7}&0
\\ 0&0&0&0\\ 0&0&0&0\end {array}
\right]
.
$$
The corresponding system is
$$
\begin{aligned}
x_1 -\frac{2}{7}x_3 +x_4&=0\\
x_2-\frac{3}{7}x_3 &=0\\
\end{aligned}.
$$
The solution of the above system is $x_1= \frac{2}{7}s-t$,$x_2=\frac{3}{7}s$,$x_3=s$, $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form
$$x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned} \frac{2}{7}s-t\\\frac{3}{7}s\\s\\ t \end{aligned}\right] = s \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right] +t \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned} \frac{2}{7} \\\frac{3}{7} \\1\\ 0 \end{aligned}\right], \left[\begin{aligned}-1\\0\\0\\ 1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $2$.