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