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AJBB

  • one year 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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  1. blockcolder
    • one year ago
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    \[A\begin{bmatrix}1\\1\\2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}; A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}; A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}\] 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
    • one year ago
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    I don't quite understand. Could you please explain?

  3. Zarkon
    • one year ago
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    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
    • one year ago
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    On it.

  5. AJBB
    • one year ago
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    Could you tell me what for, please?

  6. Zarkon
    • one year ago
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    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
    • one year ago
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    then T(x,y,z)=xT(1,0,0)+yT(0,1,0)+zT(0,0,1)

  8. AJBB
    • one year ago
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    Is this correct |dw:1379741449340:dw| ?

  9. Zarkon
    • one year ago
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    no

  10. Zarkon
    • one year ago
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    \[\left[\begin{array}{ccccc} 1 & 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ \end{array}\right] \]

  11. AJBB
    • one year ago
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    Ok. And then what should I do?

  12. Zarkon
    • one year ago
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    then T(1,0,0)=(1,0) T(0,1,0)=(1,1) T(0,0,1)=(0,1)

  13. Zarkon
    • one year ago
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    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
    • one year ago
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    Could you explain this with words please? I am not following you.

  15. Zarkon
    • one year ago
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    I just fond out where the standard basis vectors get mapped to. Once you have those the problem is easy

  16. Zarkon
    • one year ago
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    if this is confusing you can always try blockcolder's method...that will work too.

  17. AJBB
    • one year ago
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    I just don't understand this at all. I want too, I just can't seem to get it in my head.

  18. Zarkon
    • one year ago
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    do you understand where my original matrix came from?

  19. AJBB
    • one year ago
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    No, not really. I would really appreciate it if you could explain it step-by-step, please.

  20. Zarkon
    • one year ago
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    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
    • one year ago
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    look at the first row...it is just T(1,1,2)=(2,3) 1 1 2 | 2 3

  22. AJBB
    • one year ago
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    Ok.

  23. Zarkon
    • one year ago
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    same with the 2nd and 3rd rows

  24. Zarkon
    • one year ago
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    then you row reduce it....this gives us what the standard basis vectors map to

  25. AJBB
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    And then what do we do?

  28. Zarkon
    • one year ago
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    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
    • one year ago
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    then we use the linearity of T to get the answer

  30. Zarkon
    • one year ago
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    we want T(x,y,z)

  31. Zarkon
    • one year ago
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    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
    • one year ago
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    and then?

  33. Zarkon
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    but where do you get (x,0),(y,y) and (0,z)?

  36. Zarkon
    • one year ago
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    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
    • one year ago
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    T(1,0,0)=(1,0) first row of matrix 1 0 0 | 1 0

  38. AJBB
    • one year ago
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    oh, got it!

  39. Zarkon
    • one year ago
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    good

  40. AJBB
    • one year ago
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    Would it be the same process for linear transformations that for up instead of down,s ay from R2 -> R3?

  41. Zarkon
    • one year ago
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    yes

  42. AJBB
    • one year ago
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    Ok! By any chance, do you have practice problems or do you know where I could find any?

  43. Zarkon
    • one year ago
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    none typed up

  44. Zarkon
    • one year ago
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    you have a text book?

  45. AJBB
    • one year ago
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    I do, but it doesn't have linear transformations in it.

  46. Zarkon
    • one year ago
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    what book do you have?

  47. AJBB
    • one year ago
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    Found them! I was looking in the wrong chapter. Anyways, its Linear Algebra by Bernard Kolman and David R. Hill.

  48. Zarkon
    • one year ago
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    ok...I have a lot of linear algebra book...but I don't think I have that one

  49. Zarkon
    • one year ago
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    *books

  50. Zarkon
    • one year ago
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    I have like 10 of them...i'd have to check my office ;)

  51. AJBB
    • one year ago
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    Its ok. My book is in Spanish anyway. How much longer are you going to stay around?

  52. Zarkon
    • one year ago
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    not much longer...it is pretty late for me

  53. Zarkon
    • one year ago
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    2am

  54. AJBB
    • one year ago
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    Ok! Thank you so much for your help, patience and time!

  55. Zarkon
    • one year ago
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    no problem...good luck on your exam!

  56. AJBB
    • one year ago
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    Thank you!

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