If rref$([A|b])=[B|d]$, find $B$ and $d$. Let $A$ be a $3\times3$ matrix and let the complete solution of $Ax=b$ be $x=\begin{bmatrix}
4\\
0\\
0
\end{bmatrix}+$ $\alpha\begin{bmatrix}
2\\
1\\
0
\end{bmatrix}+$ $\beta\begin{bmatrix}
5\\
0\\
1
\end{bmatrix}$, where $\alpha$ and $\beta$ are any two scalars.
If rref$([A|b])=[B|d]$. Find $B$ and $d$.
The first part of this question asks for a basis of $N(A)$ and from the complete solution, it is obvious what a basis of $N(A)$ is. How does one go about using the particular and special solutions to find what the question is asking for, i.e., $B$ and $d$?
 A: $$A\boldsymbol x = \boldsymbol b$$
$$A\left(\begin{bmatrix}
4\\
0\\
0
\end{bmatrix}+\alpha\begin{bmatrix}
2\\
1\\
0
\end{bmatrix}+\beta\begin{bmatrix}
5\\
0\\
1
\end{bmatrix}\right) = \boldsymbol b$$
$$A\begin{bmatrix}
4&2&5\\
0&1&0\\
0&0&1
\end{bmatrix}\begin{bmatrix}
1\\
\alpha\\
\beta
\end{bmatrix} = \boldsymbol b$$
The solution space does not change after reducing
$$B\begin{bmatrix}
4&2&5\\
0&1&0\\
0&0&1
\end{bmatrix}\begin{bmatrix}
1\\
\alpha\\
\beta
\end{bmatrix} = \boldsymbol d$$
$$\begin{bmatrix}
k_{11}&k_{12}&k_{13}\\
0&k_{22}&k_{23}\\
0&0&k_{33}
\end{bmatrix}\begin{bmatrix}
4+2\alpha+5\beta\\
\alpha\\
\beta
\end{bmatrix} = \boldsymbol d$$
$$\begin{bmatrix}
k_{11}(4+2\alpha+5\beta)&+&k_{12}\alpha&+&k_{13}\beta\\
&&k_{22}\alpha&+&k_{23}\beta\\
&&&&k_{33}\beta
\end{bmatrix} = \begin{bmatrix}d_1\\d_2\\d_3\end{bmatrix}$$
You can then fill in the matrix entries - in particular, $\alpha,\beta$ need to cancel completely. For any $\alpha, \beta\in\mathbb R$ that multiplication needs to produce the same $\boldsymbol d$. Proceed by inspection or multiply out if you wish.
Remember $B$ is rref (the solution is unique).
A: If $v_1+\alpha v_2 + \beta v_3$ is the solution to $Ax=b$ with the vectors linearly independent, then $v_1$ and $v_2$ form a basis for the null space.  The key insight is that $A$ and $\operatorname{rref}(A)$ have the same null space.  So $\operatorname{rref}(A)$ must have two free columns, and one pivot column.  This forces there to be only one non-zero row, and it must be orthogonal to both $v_2$ and $v_3$.  This will tell you what $B$ must be.  Then, because the row reduction does not change the solution space, we must have that $d=Bv_1$.
The task, now, is to unravel this paragraph and fill in the details.
