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

Zarkon
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Could you tell me what for, please?

Zarkon
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Is this correct dw:1379741449340:dw ?

Zarkon
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Ok. And then what should I do?

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

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

AJBB
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.2same with the 2nd and 3rd rows

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

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

Zarkon
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0I do, but it doesn't have linear transformations in it.

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

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

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

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