<s>[INST] Find $A$ if solution to $Ax=b$ is given 
Solution to $Ax=b$ is $x=\begin{bmatrix}
1\\ 
0\\ 
1\\ 
0
\end{bmatrix}+\alpha_{1}\begin{bmatrix}
1\\ 
1\\ 
-1\\ 
0
\end{bmatrix}+\alpha_{2}\begin{bmatrix}
1\\ 
0\\ 
1\\ 
1\end{bmatrix}$. For $b=(1,2,1)^T$, find $A$.

Somehow, Gilbert Strang says that it is obvious that first and third column should add up to $(1,2,1)^T$, and then he says that second column is third minus first and fourth is -(first + third). I don't see how he concluded all of this, so any clarification is very welcome.
 [/INST] You have that
$$
A
\begin{bmatrix}
1+\alpha_1+\alpha_2 \\
\alpha_1 \\
1-\alpha_1+\alpha_2 \\
\alpha_2
\end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}
$$
If $A=[a_1\ a_2\  a_3\ a_4]$, this means that
$$
(1+\alpha_1+\alpha_2)a_1
+\alpha_1a_2
+(1-\alpha_1+\alpha_2)a_3
+\alpha_2 a_4
=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}=b
$$
for every $\alpha_1$ and $\alpha_2$.
In particular you have
$$
\begin{cases}
a_1+a_3=b & (\alpha_1=0, \alpha_2=0) \\[4px]
2a_1+a_2+a_3=b & (\alpha_1=1, \alpha_2=0) \\[4px]
2a_1+a_3+a_4=b & (\alpha_1=0, \alpha_2=1)
\end{cases}
$$
This is quite similar to a linear system, isn't it?
$$
\begin{bmatrix}
1 & 0 & 1 & 0 & b \\
2 & 1 & 1 & 0 & b \\
2 & 0 & 1 & 1 & b \\
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 & 0 & b \\
0 & 1 & -1 & 0 & -b \\
0 & 0 & -1 & 1 & -b \\
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 & 0 & b \\
0 & 1 & -1 & 0 & -b \\
0 & 0 & -1 & 1 & -b \\
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 & 0 & b \\
0 & 1 & -1 & 0 & -b \\
0 & 0 & 1 & -1 & b \\
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 \\
0 & 0 & 1 & -1 & b \\
\end{bmatrix}
$$
Thus $a_1=-a_4$, $a_2=a_4$, $a_3=a_4+b$. The fourth column can be anything; so the matrix $A$ is
$$
\begin{bmatrix} -x &  x &  x+b &  x \end{bmatrix}
$$
where $x$ is an arbitrary $3\times 1$ column.
</s>