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
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
Assuming that $a,b$ are scalars, it is just $\frac{ab}{a^2+b^2}I_n$
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 $(a^2I_n+b^2I_n)^{-1}(aI_n\ \ bI_n)=(a^2+b^2)^{-1}I_n(aI_n\ \ bI_n)=(a^2+b^2)^{-1}(aI_n\ \ bI_n)=\left({a\over a^2+b^2}I_n\ \ {b\over a^2+b^2}I_n\right)$