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.

- anonymous

- katieb

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- 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

I don't quite understand. Could you please explain?

- 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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## More answers

- anonymous

On it.

- anonymous

Could you tell me what for, please?

- 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

then T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)

- anonymous

Is this correct |dw:1379741449340:dw| ?

- Zarkon

no

- Zarkon

\[\left[\begin{array}{ccccc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right] \]

- anonymous

Ok. And then what should I do?

- Zarkon

then
T(1,0,0)=(1,0)
T(0,1,0)=(1,1)
T(0,0,1)=(0,1)

- 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

Could you explain this with words please? I am not following you.

- Zarkon

I just fond out where the standard basis vectors get mapped to. Once you have those the problem is easy

- Zarkon

if this is confusing you can always try blockcolder's method...that will work too.

- anonymous

I just don't understand this at all. I want too, I just can't seem to get it in my head.

- Zarkon

do you understand where my original matrix came from?

- anonymous

No, not really. I would really appreciate it if you could explain it step-by-step, please.

- 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

look at the first row...it is just
T(1,1,2)=(2,3)
1 1 2 | 2 3

- anonymous

Ok.

- Zarkon

same with the 2nd and 3rd rows

- Zarkon

then you row reduce it....this gives us what the standard basis vectors map to

- 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

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

And then what do we do?

- 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

then we use the linearity of T to get the answer

- Zarkon

we want T(x,y,z)

- 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

and then?

- 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

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

but where do you get (x,0),(y,y) and (0,z)?

- 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

T(1,0,0)=(1,0)
first row of matrix
1 0 0 | 1 0

- anonymous

oh, got it!

- Zarkon

good

- anonymous

Would it be the same process for linear transformations that for up instead of down,s ay from R2 -> R3?

- Zarkon

yes

- anonymous

Ok! By any chance, do you have practice problems or do you know where I could find any?

- Zarkon

none typed up

- Zarkon

you have a text book?

- anonymous

I do, but it doesn't have linear transformations in it.

- Zarkon

what book do you have?

- anonymous

Found them! I was looking in the wrong chapter. Anyways, its Linear Algebra by Bernard Kolman and David R. Hill.

- Zarkon

ok...I have a lot of linear algebra book...but I don't think I have that one

- Zarkon

*books

- Zarkon

I have like 10 of them...i'd have to check my office ;)

- anonymous

Its ok. My book is in Spanish anyway.
How much longer are you going to stay around?

- Zarkon

not much longer...it is pretty late for me

- Zarkon

2am

- anonymous

Ok! Thank you so much for your help, patience and time!

- Zarkon

no problem...good luck on your exam!

- anonymous

Thank you!

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