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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.
\[A\begin{bmatrix}1\\1\\2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}; A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}; A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}\] 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.
I don't quite understand. Could you please explain?
put the following matrix into rref \[\left[\begin{array}{ccccc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]\]
Could you tell me what for, please?
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
then T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)
Is this correct |dw:1379741449340:dw| ?
\[\left[\begin{array}{ccccc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right] \]
Ok. And then what should I do?
then T(1,0,0)=(1,0) T(0,1,0)=(1,1) T(0,0,1)=(0,1)
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)
Could you explain this with words please? I am not following you.
I just fond out where the standard basis vectors get mapped to. Once you have those the problem is easy
if this is confusing you can always try blockcolder's method...that will work too.
I just don't understand this at all. I want too, I just can't seem to get it in my head.
do you understand where my original matrix came from?
No, not really. I would really appreciate it if you could explain it step-by-step, please.
look at \[\left[\begin{array}{ccc|cc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]\]
look at the first row...it is just T(1,1,2)=(2,3) 1 1 2 | 2 3
same with the 2nd and 3rd rows
then you row reduce it....this gives us what the standard basis vectors map to
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.
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)
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)
then we use the linearity of T to get the answer
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)
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)
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
but where do you get (x,0),(y,y) and (0,z)?
from our rref matrix \[\left[\begin{array}{ccc|cc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right]\]
T(1,0,0)=(1,0) first row of matrix 1 0 0 | 1 0
Would it be the same process for linear transformations that for up instead of down,s ay from R2 -> R3?
Ok! By any chance, do you have practice problems or do you know where I could find any?
I do, but it doesn't have linear transformations in it.
Found them! I was looking in the wrong chapter. Anyways, its Linear Algebra by Bernard Kolman and David R. Hill.
ok...I have a lot of linear algebra book...but I don't think I have that one
I have like 10 of them...i'd have to check my office ;)
Its ok. My book is in Spanish anyway. How much longer are you going to stay around?
not much longer...it is pretty late for me
Ok! Thank you so much for your help, patience and time!
no problem...good luck on your exam!