anonymous
  • anonymous
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.
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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blockcolder
  • blockcolder
\[A\begin{bmatrix}1\\1\\2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}; A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}; A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}\] 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.
anonymous
  • anonymous
I don't quite understand. Could you please explain?
Zarkon
  • Zarkon
put the following matrix into rref \[\left[\begin{array}{ccccc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]\]

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anonymous
  • anonymous
On it.
anonymous
  • anonymous
Could you tell me what for, please?
Zarkon
  • Zarkon
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
Zarkon
  • Zarkon
then T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)
anonymous
  • anonymous
Is this correct |dw:1379741449340:dw| ?
Zarkon
  • Zarkon
no
Zarkon
  • Zarkon
\[\left[\begin{array}{ccccc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right] \]
anonymous
  • anonymous
Ok. And then what should I do?
Zarkon
  • Zarkon
then T(1,0,0)=(1,0) T(0,1,0)=(1,1) T(0,0,1)=(0,1)
Zarkon
  • Zarkon
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)
anonymous
  • anonymous
Could you explain this with words please? I am not following you.
Zarkon
  • Zarkon
I just fond out where the standard basis vectors get mapped to. Once you have those the problem is easy
Zarkon
  • Zarkon
if this is confusing you can always try blockcolder's method...that will work too.
anonymous
  • anonymous
I just don't understand this at all. I want too, I just can't seem to get it in my head.
Zarkon
  • Zarkon
do you understand where my original matrix came from?
anonymous
  • anonymous
No, not really. I would really appreciate it if you could explain it step-by-step, please.
Zarkon
  • Zarkon
look at \[\left[\begin{array}{ccc|cc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]\]
Zarkon
  • Zarkon
look at the first row...it is just T(1,1,2)=(2,3) 1 1 2 | 2 3
anonymous
  • anonymous
Ok.
Zarkon
  • Zarkon
same with the 2nd and 3rd rows
Zarkon
  • Zarkon
then you row reduce it....this gives us what the standard basis vectors map to
anonymous
  • anonymous
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.
Zarkon
  • Zarkon
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)
anonymous
  • anonymous
And then what do we do?
Zarkon
  • Zarkon
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)
Zarkon
  • Zarkon
then we use the linearity of T to get the answer
Zarkon
  • Zarkon
we want T(x,y,z)
Zarkon
  • Zarkon
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)
anonymous
  • anonymous
and then?
Zarkon
  • Zarkon
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)
Zarkon
  • Zarkon
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
anonymous
  • anonymous
but where do you get (x,0),(y,y) and (0,z)?
Zarkon
  • Zarkon
from our rref matrix \[\left[\begin{array}{ccc|cc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right]\]
Zarkon
  • Zarkon
T(1,0,0)=(1,0) first row of matrix 1 0 0 | 1 0
anonymous
  • anonymous
oh, got it!
Zarkon
  • Zarkon
good
anonymous
  • anonymous
Would it be the same process for linear transformations that for up instead of down,s ay from R2 -> R3?
Zarkon
  • Zarkon
yes
anonymous
  • anonymous
Ok! By any chance, do you have practice problems or do you know where I could find any?
Zarkon
  • Zarkon
none typed up
Zarkon
  • Zarkon
you have a text book?
anonymous
  • anonymous
I do, but it doesn't have linear transformations in it.
Zarkon
  • Zarkon
what book do you have?
anonymous
  • anonymous
Found them! I was looking in the wrong chapter. Anyways, its Linear Algebra by Bernard Kolman and David R. Hill.
Zarkon
  • Zarkon
ok...I have a lot of linear algebra book...but I don't think I have that one
Zarkon
  • Zarkon
*books
Zarkon
  • Zarkon
I have like 10 of them...i'd have to check my office ;)
anonymous
  • anonymous
Its ok. My book is in Spanish anyway. How much longer are you going to stay around?
Zarkon
  • Zarkon
not much longer...it is pretty late for me
Zarkon
  • Zarkon
2am
anonymous
  • anonymous
Ok! Thank you so much for your help, patience and time!
Zarkon
  • Zarkon
no problem...good luck on your exam!
anonymous
  • anonymous
Thank you!

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