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