# Transformation, matrices

What am I missing here?

Let $A,B$ be matrix-valued functions of $(x,y)$. Given that $$A_x-B_y+AB-BA=0$$ where the subscript denotes partial differentiation w.r.t. that variable, I wish to show that $$\alpha_x-\beta_y+\alpha\beta-\beta\alpha=0$$ where for an invertible matrix $M$, $\alpha=MAM^{-1}+M_yM^{-1}$ and $\beta=MBM^{-1}+M_xM^{-1}$

What I've done: I tried to directly substitute the $\alpha,\beta$ into the equation, giving: $$[M_xAM^{-1}+MA_xM^{-1}+MA(M^{-1})_x+M_{xy}M^{-1}+M_{y}(M^{-1})_x]-[M_yBM^{-1}+MB_yM^{-1}+MB(M^{-1})_y+M_{xy}M^{-1}+M_{x}(M^{-1})_y]+MABM^{-1}+M_yBM^{-1}+MAM^{-1}M_xM^{-1}+M_yM^{-1}M_xM^{-1}-MBAM^{-1}-M_xAM^{-1}-MBM^{-1}M_yM^{-1}-M_xM^{-1}M_yM^{-1}$$

Cancelling and using $A_x-B_y+AB-BA=0$, $$=[MA(M^{-1})_x+M_{y}(M^{-1})_x]-[MB_yM^{-1}+M_{x}(M^{-1})_y]+MAM^{-1}M_xM^{-1}+M_yM^{-1}M_xM^{-1}-MBM^{-1}M_yM^{-1}-M_xM^{-1}M_yM^{-1}$$

However, I don't see how to proceed. Any help is appreciated.

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Maybe you can use the fact that $(\Delta^{-1})_x = - \Delta^{-1} \Delta_x \Delta^{-1}$ (cf. the derivative of $f(x)=\frac{1}{x}$ is $f'(x) = -\frac{1}{x^2}$). – copper.hat Nov 22 '12 at 16:12
thank you, @copper.hat !! – Greg Nov 23 '12 at 10:22