## AJBB Group Title 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. 10 months ago 10 months ago

1. blockcolder Group Title

$A\begin{bmatrix}1\\1\\2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}; A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}; A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}$ 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.

2. AJBB Group Title

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

3. Zarkon Group Title

put the following matrix into rref $\left[\begin{array}{ccccc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]$

4. AJBB Group Title

On it.

5. AJBB Group Title

Could you tell me what for, please?

6. Zarkon Group Title

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

7. Zarkon Group Title

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

8. AJBB Group Title

Is this correct |dw:1379741449340:dw| ?

9. Zarkon Group Title

no

10. Zarkon Group Title

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

11. AJBB Group Title

Ok. And then what should I do?

12. Zarkon Group Title

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

13. Zarkon Group Title

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)

14. AJBB Group Title

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

15. Zarkon Group Title

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

16. Zarkon Group Title

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

17. AJBB Group Title

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

18. Zarkon Group Title

do you understand where my original matrix came from?

19. AJBB Group Title

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

20. Zarkon Group Title

look at $\left[\begin{array}{ccc|cc} 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 2 & 1 \\ \end{array}\right]$

21. Zarkon Group Title

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

22. AJBB Group Title

Ok.

23. Zarkon Group Title

same with the 2nd and 3rd rows

24. Zarkon Group Title

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

25. AJBB Group Title

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.

26. Zarkon Group Title

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)

27. AJBB Group Title

And then what do we do?

28. Zarkon Group Title

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)

29. Zarkon Group Title

then we use the linearity of T to get the answer

30. Zarkon Group Title

we want T(x,y,z)

31. Zarkon Group Title

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)

32. AJBB Group Title

and then?

33. Zarkon Group Title

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)

34. Zarkon Group Title

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

35. AJBB Group Title

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

36. Zarkon Group Title

from our rref matrix $\left[\begin{array}{ccc|cc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right]$

37. Zarkon Group Title

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

38. AJBB Group Title

oh, got it!

39. Zarkon Group Title

good

40. AJBB Group Title

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

41. Zarkon Group Title

yes

42. AJBB Group Title

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

43. Zarkon Group Title

none typed up

44. Zarkon Group Title

you have a text book?

45. AJBB Group Title

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

46. Zarkon Group Title

what book do you have?

47. AJBB Group Title

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

48. Zarkon Group Title

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

49. Zarkon Group Title

*books

50. Zarkon Group Title

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

51. AJBB Group Title

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

52. Zarkon Group Title

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

53. Zarkon Group Title

2am

54. AJBB Group Title

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

55. Zarkon Group Title

no problem...good luck on your exam!

56. AJBB Group Title

Thank you!