# Assume that A is row equivalent to B. Find bases for Nul A and Col A.

Assume that A is row equivalent to B. Find bases for Nul A and Col A.
$A=\left[\begin{array}{ccccc}1& 2& -5& 11& -3\\ 2& 4& -5& 15& 2\\ 1& 2& 0& 4& 5\\ 3& 6& -5& 19& -2\end{array}\right]$
$B=\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 5& -7& 8\\ 0& 0& 0& 0& -9\\ 0& 0& 0& 0& 0\end{array}\right]$

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Since B is a row echelon form of A, we see that the first, third and fifth columns of A are its pivot columns. Thus a basis fo Col A is
$\left\{\left[\begin{array}{c}1\\ 2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}-5\\ -5\\ 0\\ -5\end{array}\right],\left[\begin{array}{c}-3\\ 2\\ 5\\ -2\end{array}\right]\right\}$
To find a basis for Nul A, we find the general solution of Ax=0 in terms of the free variables. Since it is row equivalent to B we can simply get reduced row echelon form of B:

$\begin{array}{c}{R}_{1}\to {R}_{1}-5{R}_{3}\\ {R}_{2}\to {R}_{2}-\frac{8}{5}{R}_{3}\end{array}\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 1& \frac{-7}{5}& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right]$
to get: , with ${x}_{2}$ and ${x}_{4}$ free. So
$x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\\ {x}_{5}\end{array}\right]={x}_{2}\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right]+{x}_{4}\left[\begin{array}{c}-4\\ 0\\ 7/5\\ 1\\ 0\end{array}\right]$
And a basis for Nul A is
$\left\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 7/5\\ 1\\ 0\end{array}\right]\right\}$
Result: Basis for Col A is:
$\left\{\left[\begin{array}{c}1\\ 2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}-5\\ -5\\ 0\\ -5\end{array}\right],\left[\begin{array}{c}-3\\ 2\\ 5\\ -2\end{array}\right]\right\}$
Basis for Nul A is
$\left\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 7/5\\ 1\\ 0\end{array}\right]\right\}$