Answer
Basis, $B=\left\{\begin{bmatrix}1\\2\\3\\0\end{bmatrix},\begin{bmatrix}1\\0\\-1\\-1\end{bmatrix}\right\}$
The two vectors are linearly independent therefore, the dimension is 2
Work Step by Step
We are required to find the basis and state the dimension:
Let the given subspace be represented by a vector $\mathbf{\vec{z}}$ such that:
$\mathbf{\vec{z}}=\begin{bmatrix}a+b\\2a\\3a-b\\-b\end{bmatrix} $
Writing the given set in parametric form;
$\mathbf{\vec{z}}=\begin{bmatrix}a+b\\2a\\3a-b\\-b\end{bmatrix}=a\begin{bmatrix}1\\2\\3\\0\end{bmatrix}+b\begin{bmatrix}1\\0\\-1\\-1\end{bmatrix}$
The given subspace is a linear combination of two vectors
$\begin{bmatrix}1\\2\\3\\0\end{bmatrix},\begin{bmatrix}1\\0\\-1\\-1\end{bmatrix}$
Since the two vectors are linearly independent, they form the basis for the set.
$B=\left\{\begin{bmatrix}1\\2\\3\\0\end{bmatrix},\begin{bmatrix}1\\0\\-1\\-1\end{bmatrix}\right\}$
The two vectors are linearly independent therefore, the dimension is 2
