Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned}\frac{4}{3} \\-\frac{3}{2} \\-1\\1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $1$.
Work Step by Step
The coefficient matrix is given by
$$
\left[ \begin {array}{cccc} 9&-4&-2&-20\\ 12&-6&-4&
-29\\ 3&-2&0&-7\\ 3&-2&-1&-8
\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{cccc} 1&0&0&-\frac{4}{3}\\ 0&1&0&\frac{3}{2}
\\ 0&0&1&1\\ 0&0&0&0\end {array}
\right]
.
$$
The corresponding system is
$$
\begin{aligned}
x_1 - \frac{4}{3}x_4&=0\\
x_2+\frac{3}{2}x_4&=0\\
x_3+x_4&=0\\
\end{aligned}.
$$
The solution of the above system is $x_1=\frac{4}{3}t$,$x_2=-\frac{3}{2}t$,$x_3=-t$, $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form
$$x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}\frac{4}{3}t\\-\frac{3}{2}t\\-t\\t \end{aligned}\right] = t\left[\begin{aligned}\frac{4}{3} \\-\frac{3}{2} \\-1\\1 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned}\frac{4}{3} \\-\frac{3}{2} \\-1\\1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $1$.
