A= \begin{bmatrix}4& 1& 11 \\ 1& 3& 11 \\ 0& 2& 6 \\1& 1& 5\end{bmatrix}
B= \begin{bmatrix}2 & -3 & 6 & -5 \\-2 & 2 & -5 & 6 \end{bmatrix}
Row reduce
\begin{bmatrix}4& 1& 11 & a\\ 1& 3& 11 & b \\ 0& 2& 6 & c\\1& 1& 5 & d\end{bmatrix}
and you will get something like
\begin{bmatrix}1& 1& 5 & d\\ 0& 2& 6 & c \\ 0& 0 & 0 & b-d-c\\0& 0 & 0 & a- 4d+ \frac{3}{2}c\end{bmatrix}
In order that that be solvable we must have b- d- c= 0 and a- 4d+ (3/2)c= 0. Those lead to a= 4d- (3/2)c and b= c+ d so the space of all such b is spanned by <4, 1, 0 1> and <-3, 2, 0, 1> (by taking d=1, c=0 and then d=0, c= 2).
If you row reduce
\begin{bmatrix}2 & -3 & 6 & -5 & 0 \\-2 & 2 & -5 & 6 & 0 \end{bmatrix}
you get something like
\begin{bmatrix} 2 & - 3 & 6 & -5 \\0 & -1 & 1 & 1\end{bmatrix}
The Null space of B is given by <a, b, c, d> with 2a- 3b+ 6c- 5d= 0 and -b+ c+ d= 0.
Show that the solutions to those equations are exactly the space above.
R= \begin{bmatrix}1 & -1 & 2 & -2 \\ -1 & 2 & -3 & 2 \\ 2 & -3 & 5 & -5\\ -2 & 2 & -5 & 6\end{bmatrix}
That is a square matrix so if you can show it has an inverse, all you have to do is multiply, on the left, on both sides of RAx= Rb by the inverse of R!
