Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned}0\\0 \\0 \end{aligned}\right].$$
(b) The dimension of the solution space is $0$.
Work Step by Step
The coefficient matrix is given by
$$
\left[ \begin {array}{ccc} 4&-1&2\\ 2&3&-1
\\ 3&1&1\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{ccc} 1&0&0\\ 0&1&0
\\ 0&0&1\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned} x &=0\\
y &=0\\
z&=0
\end{aligned}.
$$
The solution of the above system is $x=0$,$y=0$, $z=0$. This means that the solution space of $Ax = 0 $ consists of the zero vector
$$x= \left[\begin{aligned} x\\ y\\z \end{aligned}\right]= \left[\begin{aligned}0\\0 \\0 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned}0\\0 \\0 \end{aligned}\right].$$
(b) The dimension of the solution space is $0$.
