Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (anonymous):

determine if function T: R^2 --> R^2 are one to one or onto T(x,y)=(2x,y)

OpenStudy (anonymous):

is there any more info on this?

OpenStudy (anonymous):

nope, that is it..

OpenStudy (anonymous):

what is the name of the section this is out of form your textbook?

OpenStudy (anonymous):

geometry of maps from R2 to R2, vector calc.

OpenStudy (haseeb96):

this function is one one onto

OpenStudy (anonymous):

oooohh.... ok, when you say R^2 you mean mapping from R2 to R2. its easier than what were making it out to be

OpenStudy (anonymous):

what do i do to know if it is one to one or onto.

OpenStudy (anonymous):

yeah like 2d

OpenStudy (anonymous):

does your textbook have a section on "one to one" vectors?

OpenStudy (amistre64):

let \(\vec x\) = (x,y) T(\(\vec x\)) = A\(\vec x\) \[\begin{pmatrix}2&0\\0&1\end{pmatrix}\binom{x}{y}=\binom{2x}{y}\]

OpenStudy (anonymous):

okay so you put a derivative in a determinant ?

OpenStudy (amistre64):

not a derivative :) i made a matrix A that transforms any given input (x,y) and produces an output (2x,y)

OpenStudy (amistre64):

the mapping is one to one if every input has a unique output. can we create an inverse of this? so that for every 2x,y, we can get back to x,y ?

OpenStudy (anonymous):

not sure ?

OpenStudy (amistre64):

\[\begin{pmatrix}\frac 12&0\\0&1\end{pmatrix}\binom{2x}{y}=\binom{x}{y}\]

OpenStudy (amistre64):

if you are not familiar with matrixes ... then none of this will make sense

OpenStudy (anonymous):

...none of it makes sense. haha. the book is saying that if it is one to one if T(u,v) = T(u',v')

OpenStudy (amistre64):

im not sure that ' denotes a derivative in this case

OpenStudy (amistre64):

|dw:1383594049088:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
luvnickk: how do we delete accounts on here?
15 hours ago 2 Replies 0 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!