Answer
Basis, $B=\left\{\begin{bmatrix}0\\1\\0\\1\end{bmatrix},\begin{bmatrix}0\\-1\\1\\2\end{bmatrix},\begin{bmatrix}2\\0\\-3\\0\end{bmatrix}\right\}$
As there are 3 pivot columns the dimension is 3
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}2c\\a-b\\b-3c\\a+2b\end{bmatrix} $
Writing the given set in parametric form;
$\mathbf{\vec{z}}=\begin{bmatrix}2c\\a-b\\b-3c\\a+2b\end{bmatrix}=a\begin{bmatrix}0\\1\\0\\1\end{bmatrix}+b\begin{bmatrix}0\\-1\\1\\2\end{bmatrix}+c\begin{bmatrix}2\\0\\-3\\0\end{bmatrix}$
The given subspace is a linear combination of three vectors,
$\begin{bmatrix}0\\1\\0\\1\end{bmatrix},\begin{bmatrix}0\\-1\\1\\2\end{bmatrix}and\begin{bmatrix}2\\0\\-3\\0\end{bmatrix}$
from the vectors let us form an augmented matrix $\mathbf{M}$ and row reduce it to echelon form ,
$\mathbf{M}=\begin{bmatrix}0&0&2\\1&-1&0\\0&1&-3\\1&2&0\end{bmatrix}\sim\begin{bmatrix}0&0&2\\1&-1&0\\0&1&-3\\0&0&1\end{bmatrix}$
We have 3 pivot elements hence the vectors are linearly independent, and the 3 vectors form the basis for the set.
$B=\left\{\begin{bmatrix}0\\1\\0\\1\end{bmatrix},\begin{bmatrix}0\\-1\\1\\2\end{bmatrix},\begin{bmatrix}2\\0\\-3\\0\end{bmatrix}\right\}$
As there are 3 pivot columns the dimension is 3
