So I am trying to solve some problems from Hughston and Tod's Introduction to General Relativity. I need help with the following system of equations:
$$aX_i+bY_i=P_i$$ $$\epsilon_{ijk} X_jY_k=Q_i$$
From the first equation we get: $$Y_i=P_i/b-aX_i/b$$
plugging into the second equation yields: $$\epsilon_{ijk} X_j P_k=b Q_i$$ Now I am stuck and not sure how to solve it. Help would be appreciated. Please if possible use index-notation.
In vector form we can write the equations as $$a\underline{x}+b\underline{y}=\underline{p}$$ and $$\underline{x}\times\underline{y}=\underline{q}.$$ Therefore the vectors $\underline{x}$, $\underline{y}$ and $\underline{p}$ are co-planar, and $\underline{q}$ is perpendicular to this plane. It follows that $\underline{p}\times\underline{q}$ is parallel to this plane and perpendicular to $\underline{p}$.
Therefore the vectors $\underline{x}$ and $\underline{y}$ can be expressed as a linear combination of $\underline{p}$ and $\underline{p}\times\underline{q}$.
If we therefore write $$\underline{x}=A\underline{p}+B\underline{p}\times\underline{q}$$ and $$\underline{y}=C\underline{p}+D\underline{p}\times\underline{q}$$ We can substitute these expressions into the given equations.
From the first we obtain $$aA+bC=1$$ and $$aB+bD=0$$
Calculating $\underline{x}\times\underline{y}$ using the vector triple product formula, gives $$(AD-BC)\left[\underline{p}(p \cdot q)-\underline{q}p^2\right]=\underline{q}$$
This leads to a third equation $$AD-BC=-\frac{1}{P^2}$$
Solving these equations, at least in part, determines that $$B=\frac{b}{p^2}$$ and $$D=-\frac{a}{p^2}$$ whereas $A$ and $C$ are arbitrary and are chosen to satisfy $$aA+bC=1$$