discovered that ${M}^{-1}$ is the opposite transformation.

What makes it true?

Kristen Stokes
2022-07-10
Answered

Let $M$ be a transformation matrix $B\to {B}^{\prime}$

discovered that ${M}^{-1}$ is the opposite transformation.

What makes it true?

discovered that ${M}^{-1}$ is the opposite transformation.

What makes it true?

You can still ask an expert for help

Marisol Morton

Answered 2022-07-11
Author has **13** answers

If $X$ is a matrix of a vector $x$ in the basis $B$ then its matrix $X\prime $ in $B\prime $ is

${X}^{\prime}=MX$

but in this case we have

$X={M}^{-1}{X}^{\prime}$

hence ${M}^{-1}$ is the transformation matrix ${B}^{\prime}\to B$

${X}^{\prime}=MX$

but in this case we have

$X={M}^{-1}{X}^{\prime}$

hence ${M}^{-1}$ is the transformation matrix ${B}^{\prime}\to B$

Ellen Chang

Answered 2022-07-12
Author has **5** answers

I'm not sure what "opposite transformations" are, but they sound like inverse transformations and that's the definition of ${M}^{-1}$'s relationship to $M$

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The inverse, ${A}^{-1}$ is ?

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where $p\in ]0,1[$ and $x>1$. I would like to have a solution $x$ in terms of $p$ for each fixed $p$ in the specified interval. Of course it's possible that no analytical form of the solution exists. If so, I'd also be happy to hear an argument why that is the case. However, for some values of p the solution is 'nice', for example for $p=\frac{1}{2}$ it's easy to compute that $x=8$ and for $p=\frac{1}{3}$ I obtained $x=\frac{9}{16}(5\sqrt{3}+\sqrt{11+64\sqrt{3}})$

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$(1+x{)}^{1-p}+{p}^{\frac{p}{1-p}}x(p-1)-1=0$

where $p\in ]0,1[$ and $x>1$. I would like to have a solution $x$ in terms of $p$ for each fixed $p$ in the specified interval. Of course it's possible that no analytical form of the solution exists. If so, I'd also be happy to hear an argument why that is the case. However, for some values of p the solution is 'nice', for example for $p=\frac{1}{2}$ it's easy to compute that $x=8$ and for $p=\frac{1}{3}$ I obtained $x=\frac{9}{16}(5\sqrt{3}+\sqrt{11+64\sqrt{3}})$

Any kind of help is greatly appreciated!

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