I am currently looking at the question:
Let $$A =\begin{bmatrix} 1& −1& 3& 1& 2\\ 4& −4& 12& 6& 0\\ −3& 3& −9& −4& −2\end{bmatrix}$$
By bringing the matrix A into row echelon form, find bases for row(A), col(A) and N(A). Determine the rank and nullity of A, and verify that the Rank-Nullity Theorem holds for the above matrix A.
I have attained that matrix $A$ in row echelon form is
$$ \begin{bmatrix}1 &-1& 3& 1& 2\\ 0 & 0& 0& 1& -4\\ 0 & 0& 0& 0& 0\end{bmatrix}$$
Do I have to continue where I put the matrix into reduced row echelon form? Furthermore, I understand how to find $\text{row}(A)$, $\text{col}(A)$ as well as $N(A)$, but how would I find the basis of these? I am struggling to understand the concept of basis and its use
We don’t need to proceed further, note also that
a basis for $col(A)$ is given be the first and fourth vectors of the original matrix (corresponding to pivot columns in RREF)
a basis for $row(A)$ is given by the two rows in the RREF
to find the null space solve the system $Ax=0$ using A in RREF; since it has dimension 3 the basis is made of three vectors