Solve for X $$2X+X^t=B$$ Where $X,B, $are matrices, and $ X^t$ stands for transpose. I was trying to work with indices but it doesn't seems to work...
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Note taking the transpose of the equation $2X+ X^T = B$, we get that $$2X^T + X = B^T$$ Solving which gives us $$2(B-2X) + X = B^T \implies X = \dfrac{2B-B^T}3 \text{ and } X^T = \dfrac{2B^T-B}3$$
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Apply transposition to your equation. Then you get two linear equations with unknowns $X$ and $X^t$. Solve this system.
Norbert
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$$(2X+X^t)^t=B^t\implies 2X^t+X=B^t\implies X^t=(B^t-X)/2\implies$$
$$\implies2X+(B^t-X)/2=B\implies 2X-X/2=B-B^t/2\implies$$ $$\implies 3X/2= B-B^t/2\implies X=(2/3)B -(1/3)B^t$$
Integral
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