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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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How come the second bracket is $1\times2$? The identity matrix, by definition, is a square matrix. So, your two bracketed terms, as well as their products, are square matrices. –  user1551 Dec 2 '12 at 23:45
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I think $(aI_nbI_n)$ is meant to be $(aI_n\ bI_n)$, a matrix with $n$ rows and $2n$ columns. –  Gerry Myerson Dec 3 '12 at 10:28
    
I see. That makes sense, but we still need the OP to confirm it. –  user1551 Dec 3 '12 at 12:16

2 Answers 2

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 $$(a^2I_n+b^2I_n)^{-1}(aI_n\ \ bI_n)=(a^2+b^2)^{-1}I_n(aI_n\ \ bI_n)=(a^2+b^2)^{-1}(aI_n\ \ bI_n)=\left({a\over a^2+b^2}I_n\ \ {b\over a^2+b^2}I_n\right)$$

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