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AJBB
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T:R3 > R2 is a linear transformation where T(1,1,2)=(2,3) T(1,0,1)=(1,1) and T(1,1,0)=(2,1) Determine T(X,Y,Z) and prove that T(X,Y,Z) is a linear transformation.
 10 months ago
 10 months ago
AJBB Group Title
T:R3 > R2 is a linear transformation where T(1,1,2)=(2,3) T(1,0,1)=(1,1) and T(1,1,0)=(2,1) Determine T(X,Y,Z) and prove that T(X,Y,Z) is a linear transformation.
 10 months ago
 10 months ago

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blockcolder Group TitleBest ResponseYou've already chosen the best response.0
\[A\begin{bmatrix}1\\1\\2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}; A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}; A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}\] If you let \[A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ \end{bmatrix}\] Then you will get two system of equations so that you can solve for the entries of A.
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
I don't quite understand. Could you please explain?
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
put the following matrix into rref \[\left[\begin{array}{ccccc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]\]
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Could you tell me what for, please?
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
it is a way of finding the answer...it will give you what T(1,0,0) T(0,1,0) and T(0,0,1) are
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
then T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Is this correct dw:1379741449340:dw ?
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[\left[\begin{array}{ccccc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right] \]
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Ok. And then what should I do?
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
then T(1,0,0)=(1,0) T(0,1,0)=(1,1) T(0,0,1)=(0,1)
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1) =x(1,0)+y(1,1)+z(0,1)=(x+y,y+z)
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Could you explain this with words please? I am not following you.
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
I just fond out where the standard basis vectors get mapped to. Once you have those the problem is easy
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
if this is confusing you can always try blockcolder's method...that will work too.
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
I just don't understand this at all. I want too, I just can't seem to get it in my head.
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
do you understand where my original matrix came from?
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
No, not really. I would really appreciate it if you could explain it stepbystep, please.
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
look at \[\left[\begin{array}{ccccc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]\]
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
look at the first row...it is just T(1,1,2)=(2,3) 1 1 2  2 3
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
same with the 2nd and 3rd rows
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
then you row reduce it....this gives us what the standard basis vectors map to
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
What is a standard basis map and what is it for? I'm sorry for all my questions. I just saw this yesterday in class and my test is tomorrow.
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
we find the standard basis...(1,0,0), (0,1,0) and (0,0,1) and what their image is under T is ie the values of T(1,0,0),T(0,1,0) and T(0,0,1)
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
And then what do we do?
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
the wording i'm using mean the following T(1,1,2)=(2,3) is the same as saying T maps (1,1,2) to (2,3)
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
then we use the linearity of T to get the answer
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
we want T(x,y,z)
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
T[(x,y,z)]=T[(x,0,0)+(0,y,0)+(0,0,z)] =T(x,0,0)+T(0,y,0)+T(0,0,z) =xT(1,0,0)+yT(0,1,0)+zT(0,0,1)
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
T[(x,y,z)]=T[(x,0,0)+(0,y,0)+(0,0,z)] =T(x,0,0)+T(0,y,0)+T(0,0,z) =xT(1,0,0)+yT(0,1,0)+zT(0,0,1) =x(1,0)+y(1,1)+z(0,1) =(x,0)+(y,y)+(0,z) =(x+y,y+z)
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
so T(x,y,z)=(x+y,y+z) if we plug in our original vectors T(1,1,2)=(1+1,1+2)=(2,3) T(1,0,1)=(1+0,0+1)=(1,1) T(1,1,0)=(1+1,1+0)=(2,1) so it looks like we are correct
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
but where do you get (x,0),(y,y) and (0,z)?
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
from our rref matrix \[\left[\begin{array}{ccccc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right]\]
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
T(1,0,0)=(1,0) first row of matrix 1 0 0  1 0
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Would it be the same process for linear transformations that for up instead of down,s ay from R2 > R3?
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Ok! By any chance, do you have practice problems or do you know where I could find any?
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
none typed up
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
you have a text book?
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
I do, but it doesn't have linear transformations in it.
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
what book do you have?
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Found them! I was looking in the wrong chapter. Anyways, its Linear Algebra by Bernard Kolman and David R. Hill.
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
ok...I have a lot of linear algebra book...but I don't think I have that one
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
I have like 10 of them...i'd have to check my office ;)
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Its ok. My book is in Spanish anyway. How much longer are you going to stay around?
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
not much longer...it is pretty late for me
 10 months ago

AJBB Group TitleBest ResponseYou've already chosen the best response.0
Ok! Thank you so much for your help, patience and time!
 10 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
no problem...good luck on your exam!
 10 months ago
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