Answer
Basis, $B=\left\{\begin{bmatrix}3\\1\\0\\0\end{bmatrix},\begin{bmatrix}-1\\0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\right\}$
The three vectors are linearly independent and are non-zero therefore, the dimension is 3
Work Step by Step
We are given the subspace;
$(a,b,c,d): a-3b+c=0.$
We are required to find a basis and dimension.
we can rewrite the equation as,
$a=3b-c$
Let the given subspace be represented in vector form by y:
$y=(3b-c,b,c,d)$
$y=(3b,b,0,0) + (-c,0,c,0) + (0,0,0,d)$
$y =b(3,1,0,0) +c(-1,0,1,0) +d(0,0,0,1)$
Such that;
$Y=\left\{b\begin{bmatrix}3\\1\\0\\0\end{bmatrix},c\begin{bmatrix}-1\\0\\1\\0\end{bmatrix},d\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\right\}$
Forming the augmented matrix A and row reducing to echelon form;
$A=\begin{bmatrix}3&-1&0\\1&0&0\\0&1&0\\0&0&1\end{bmatrix}\sim\begin{bmatrix}3&-1&0\\0&\frac{1}{3}&0\\0&0&0\\0&0&1\end{bmatrix}$
Since the vectors are linearly independent, they form the basis for the set.
$B=\left\{\begin{bmatrix}3\\1\\0\\0\end{bmatrix},\begin{bmatrix}-1\\0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\right\}$
The three vectors are linearly independent and are non-zero therefore, the dimension is 3
