Finding basis from coordinate vectors Find the basis of the plane 2x - 3y + 4z = 0 such that coordinates of the vector (1, 2, 1) are (2,3) wrt this basis. 
In this question we first find another basis A = {(3,2,0), (-2,0,1)}. Then the coordinates of v= (1,2,1) with respect to A which come out to be (1,1). We are already given the vector coordinates of v with respect to basis B ie (2,3). So how to find the basis B from here?
 A: You need to start by saying, let $\mathcal{B} = \{ v, w \}$ be a basis of the plane $2x_1 - 3x_2 + 4x_3 = 0$ such that $(1,2,1)^T_{\mathcal{B}} = (2,3)^T$. Then, what this means is that $2v + 3w = (1,2,1)^T$. That is,
if $v = (v_1,v_2,v_3)^T$ and $w = (w_1,w_2,w_3)^T$, then we have
$$
\begin{align*}
2v_1 + 3w_1 &= 1\\
2v_2 + 3w_2 &= 2\\
2v_3 + 3w_3 &= 1
\end{align*}
$$
and
$$
\begin{align*}
2v_1 -3v_2 + 4v_3 &= 0\\
2w_1 -3w_2 + 4w_3 &= 0.
\end{align*}
$$
Now, the last two equations have the general solution
$$
(v_1,v_2,v_3)^T = \left( \frac{3t_1 - 4t_2}{2},t_1,t_2 \right)^T\\
(w_1,w_2,w_3)^T = \left( \frac{3s_1 - 4s_2}{2},s_1,s_2 \right)^T
$$
where $t_1,t_2,s_1,s_2 \in \mathbb{R}$. Substituting this into the first set of equations, we get
$$
\begin{alignat*}{10}
3t_1 & {}-{} & 4t_2 & {}+{} & 3s_1 & {}-{} & 4s_2 & {}={} & 1 & \\
2t_1 &       &      & {}+{} & 3s_1 &       &      & {}={} & 2 & \\
     &       & 2t_2 &       &      & {}+{} & 3s_2 & {}={} & 1 &.
\end{alignat*}
$$
This system of $3$ linear equations in $4$ unknowns can be simplified into
$$
\begin{alignat*}{10}
t_1 &  &  &  &  & {}+{} & 2s_2 & {}={} & 1 & \\
2t_1 &       &      & {}+{} & 3s_1 &       &      & {}={} & 2 & \\
     & \phantom{+}      & 2t_2 &       &      & {}+{} & 3s_2 & {}={} & 1 &.
\end{alignat*}
$$
Clearly, for any choice of $s_2$, we have a unique choice of $t_1$, $t_2$ and $s_1$ that satisfies these equations. So, let $s_2 = 1$. Then, we get $t_1 = -1$, $t_2 = -1$ and $s_1 = 4/3$. So, we get
$$
\begin{align*}
v &= \left( \frac{1}{2},-1,-1 \right)^T\\
w &= \left( 0, \frac{4}{3}, 1 \right)^T.
\end{align*}
$$
By construction, $v$ and $w$ lie on the given plane and $2v + 3w = (1,2,1)^T$.
We can also verify that $\{ v, w\}$ is indeed a linearly independent set, so this set forms a basis of the given plane: if $c,d \in \mathbb{R}$ such that
$$
0 = (0,0,0)^T = c v + d w = \left( \frac{c}{2}, - c + \frac{4d}{3}, -c + d \right)^T
$$
then we get $c=0$ by comparing the first component, and substituting this into the third component and comparing, we get $d=0$. Hence, $\{ v,w \}$ is a linearly independent set.
