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.

  • $\begingroup$ "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 '19 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 '19 at 7:21
  • $\begingroup$ (The above assuming of course my idea is correct. It's just a guess.) $\endgroup$ – Eevee Trainer Mar 8 '19 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 '19 at 7:50
  • $\begingroup$ Thanks for your replies! $\endgroup$ – Dalton3000 Mar 8 '19 at 8:35

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,$$f^{-1}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}\frac{1}{4} (-a+b+c)\\\frac1{12}(3a+b+c)\\\frac14(3a+b-3c)\end{pmatrix}.$$

  • $\begingroup$ Makes sense, thank you! $\endgroup$ – Dalton3000 Mar 8 '19 at 8:35
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Mar 8 '19 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.