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