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 one year ago
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.
 one year ago
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.

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blockcolder
 one year ago
Best 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.

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0I don't quite understand. Could you please explain?

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

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0Could you tell me what for, please?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2it 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

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2then T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0Is this correct dw:1379741449340:dw ?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2\[\left[\begin{array}{ccccc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right] \]

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0Ok. And then what should I do?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2then T(1,0,0)=(1,0) T(0,1,0)=(1,1) T(0,0,1)=(0,1)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2T(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)

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0Could you explain this with words please? I am not following you.

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

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2if this is confusing you can always try blockcolder's method...that will work too.

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

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2do you understand where my original matrix came from?

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0No, not really. I would really appreciate it if you could explain it stepbystep, please.

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

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2look at the first row...it is just T(1,1,2)=(2,3) 1 1 2  2 3

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2same with the 2nd and 3rd rows

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2then you row reduce it....this gives us what the standard basis vectors map to

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0What 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.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2we 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)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2the 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)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2then we use the linearity of T to get the answer

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2T[(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)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2T[(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)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2so 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

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0but where do you get (x,0),(y,y) and (0,z)?

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

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2T(1,0,0)=(1,0) first row of matrix 1 0 0  1 0

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

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

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0I do, but it doesn't have linear transformations in it.

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

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

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

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

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2not much longer...it is pretty late for me

AJBB
 one year ago
Best ResponseYou've already chosen the best response.0Ok! Thank you so much for your help, patience and time!

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.2no problem...good luck on your exam!
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