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Let $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. Taking the scalar product with $\mathbf{b}\times \mathbf{c}$ in the first expansion kills its other two terms. Taking the scalar product with $\mathbf{a}$ in the second expansion does the same. Thus the two requested coordinates areThe scalar triple product condition guarantees linear independence and the existence of these reciprocal bases. The second coordinate is not obtained by merely normalizing the first basis vector: the two expansions use different bases.

The two affine lines meet at $(a,b)=(2,1)$. Translating their intersection to the origin, $X=x-2$, $Y=y-1$, removes both constant terms and gives the homogeneous first-order differential equationOn a patch with $X\ne 0$, put $Y=uX$. Then $Xdu/dX=(u^2+4)/(2-u)$, and separation of variables givesThus one local real implicit form isThe arctangent branch must be continued on the chosen solution patch. More geometrically, set $r=\sqrt{4X^2+Y^2}$ and take a continuous local argument $\theta =\arg (2X+iY)$; then $\theta -\ln r=C$. This form extends through $X=0$ where the original slope is defined. The intersection itself is excluded because the original right-hand side is $0/0$ there, and points with $2X-Y=0$ do not define a finite derivative. No real constant-$u$ solution was lost, because $u^2+4$ has no real zero. Translating the intersection works precisely because the two affine linear forms then become homogeneous linear forms.

Use the Characteristic polynomials of AB and BA, which also account for the zero eigenvalue and algebraic multiplicities. The block determinantequals both $\det (I+tAB)$ and $\det (I+tBA)$, by taking the two Schur complements. For $\lambda \ne 0$, set $t=-1/\lambda$ and multiply by ${\lambda }^n$ to obtainBoth sides are polynomials, so agreement away from zero proves equality there as well. Their roots, with multiplicity, are the same. The familiar eigenvector argument $ABv=\mu v\Rightarrow BA(Bv)=\mu Bv$ works directly only when $\mu \ne 0$, because then $Bv\ne 0$; it should not be used to discard the zero case.

Subtracting twice the first equation from the second gives $(\lambda +2)z=0$. The third equation gives $x=1+\lambda y$, and substitution in the first then gives $z=(\lambda +1)y$. Consequently $(\lambda +2)(\lambda +1)y=0$. The complete real solution set of this linear system isThere is no parameter value allowing both $x<0$ and $y<0$. In either exceptional family, $y=s<0$ forces $x>0$; in the remaining case, $y=0$. The exceptional values must be handled before dividing by their factors.