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Let $A$ be an invertible $2\times2$ matrix and $Y,Z$ be $2\times1$ matrices, all with real elements. Then clearly $x=Y^{t}AZ$ is a real number, where $Y^{t}$ denotes the transpose of $Y$. Assuming that $x$ is non-zero, is there a way to express $1/x$ in terms of $A,Y$ and $Z$?

Alternatively, is there a way to simplify the following expression: $$AZ\left( 1+Y^{t}AZ\right) \left( Y^{t}AY\right) ^{-1}$$

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Maybe I am missing something... but why can't you just do $x^{-1}=(\mathbf{Y}^\prime\mathbf{AZ})^{-1}$? – hejseb May 5 '13 at 10:33
Even when $A$ is the identity matrix, you appear to be asking for a way to express $(a_1 b_1 + a_2 b_2)^{-1}$ in terms of $Y$ and $Z$, where $Y^t = [a_1, a_2]$ and $Z^t = [b_1, b_2]$. I'm not sure whether there is a sensible formula here. – Andreas Caranti May 5 '13 at 11:50
@AndreasCaranti: Even a negative answer is still an answer. – digital-Ink May 5 '13 at 13:36

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