I have an exercise to find two basis to the null space of the following matrix:
\begin{bmatrix} 1 & 2 & 1 &1 \\[0.3em] 1 & 2 & 2 &-1 \\[0.3em] 2 & 5 & 0 &6 \end{bmatrix}
I'm going to use the fact that $N(A)= N(rref(A))$ ($rref(A)$ is the reduced row echelon form of A) and my basis will be:
- the pivot columns in $rref(A)$
- the corresponding columns in the original matrix $A$.
Ok, however by doing this I ended up a question:
When I get to:
\begin{bmatrix} 1 & 2 & 1 &1 \\[0.3em] 0 & 0 & 1 &-2 \\[0.3em] 0 & 1 & -2 &4 \end{bmatrix}
Should I exchange row 2 with row 3 to have a pivot? Also, does the operation of "exchanging rows" changes the fact that we can choose the equivalent columns in the original matrix as a basis of the null space of the matrix (tell me if I'm not being clear)
Thanks!