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I need to find the the inverse of the isomorphic function

$f:\Bbb R^3 \rightarrow \Bbb R^3$ given by $\begin{pmatrix}a\\b\\c\end{pmatrix} \rightarrow \begin{pmatrix}3b-a\\3a+c\\3b-c\end{pmatrix}$

Honestly, I'm not quite sure how to do this. I know you can find the inverse of a matrix using row operations, but I have not covered this yet so there should be an alternative way to find the inverse function. Any suggestions would be appreciated.

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  • $\begingroup$ "I know you can find the inverse of a matrix using row operations" -- That's almost certainly not going to be relevant here, for what it's worth; matrices are invertible only if they're square (unless there's something I'm overlooking or some bit of obscure knowledge I'm unaware of). Not to say I know off-hand how one would solve this, but I can at least eliminate that much with some amount of certainty. $\endgroup$ – Eevee Trainer Mar 8 at 7:19
  • $\begingroup$ ... Though I wonder. Perhaps you could construct a matrix such that, when you multiply it on the $( a b c )$ vector, you get the output you see on the right? Then you would have a matrix equation, $Ax = b$ where $x$ is $(a b c)$ and $b$ is the image under $f$. Then $x = A^{-1}b$ would give you your inverse mapping - just gotta do the calculation. $\endgroup$ – Eevee Trainer Mar 8 at 7:21
  • $\begingroup$ (The above assuming of course my idea is correct. It's just a guess.) $\endgroup$ – Eevee Trainer Mar 8 at 7:22
  • $\begingroup$ $f$ is linear and is an endomorphism of $\mathbb{R}^3$, then it admits a representation as a square matrix. @EeveeTrainer $\endgroup$ – nicomezi Mar 8 at 7:50
  • $\begingroup$ Thanks for your replies! $\endgroup$ – Dalton3000 Mar 8 at 8:35
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Solve the system$$\left\{\begin{array}{l}3b-a=x\\3a+c=y\\3b-c=z.\end{array}\right.$$You will get$$a=\frac{1}{4} (-x+y+z),\ b=\frac1{12}(3x+y+z)\text{ and }c=\frac14(3x+y-3 z).$$So,$$f^{-1}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}\frac{1}{4} (-a+b+c)\\\frac1{12}(3a+b+c)\\\frac14(3a+b-3c)\end{pmatrix}.$$

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  • $\begingroup$ Makes sense, thank you! $\endgroup$ – Dalton3000 Mar 8 at 8:35
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Mar 8 at 8:49

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