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I would like to count the below :

$(a^{2}I_{n} + b^{2}I_{n})^{-1}$ * $(aI_{n} bI_{n})$ = ?

Any idea? Note the second bracket is a matrix (1x2) and $I_{n}$ is an identity matrix.

Thanks in advance

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    I see. That makes sense, but we still need the OP to confirm it.2012-12-03

2 Answers 2

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Assuming that $a,b$ are scalars, it is just $\frac{ab}{a^2+b^2}I_n$

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The question is not entirely clear, but here goes. First, $a^2I_n+b^2I_n=(a^2+b^2)I_n$ It follows that $(a^2I_n+b^2I_n)^{-1}=(a^2+b^2)^{-1}I_n$ Now I'm going to assume that $(aI_nbI_n)$ is actually the $n$-by-$2n$ matrix $(aI_n\ \ bI_n)$. Then $(a^2I_n+b^2I_n)^{-1}(aI_n\ \ bI_n)=(a^2+b^2)^{-1}I_n(aI_n\ \ bI_n)=(a^2+b^2)^{-1}(aI_n\ \ bI_n)=\left({a\over a^2+b^2}I_n\ \ {b\over a^2+b^2}I_n\right)$