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KonradZuse

  • 3 years ago

4.10 #1. In Exercise 1 let T[a] and T[b] be the operators whose standard matrices are given. Find the standard matrices for T[a] o T[b] and T[b] o T[a]

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  1. KonradZuse
    • 3 years ago
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    A =\[\left[\begin{matrix}1 & -2 & 0 \\ 4 & 1 & -3 \\ 5 & 2 & 4\end{matrix}\right]\] B = \[\left[\begin{matrix}2 & -3 & 3 \\ 5 & 0 & 1 \\ 6 & 1 & 7\end{matrix}\right]\] @satellite73

  2. UnkleRhaukus
    • 3 years ago
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    what do you mean by the little circle o

  3. KonradZuse
    • 3 years ago
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    it's just what the book says, let me screen shot.

  4. UnkleRhaukus
    • 3 years ago
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    what does it mean

  5. KonradZuse
    • 3 years ago
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    I believe it says the composite of linear transformations.

  6. KonradZuse
    • 3 years ago
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  7. KonradZuse
    • 3 years ago
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    http://tutorial.math.lamar.edu/Classes/LinAlg/LinearTransformations.aspx This has it at the bottom as well, it's not really explaining how we get to an answer... I see from the previous section 4.9 that we use rotations and such but.....

  8. KonradZuse
    • 3 years ago
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    Save me :).

  9. KonradZuse
    • 3 years ago
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    It gives me the answer already, but I want to know why that's the answer and how to get it...

  10. UnkleRhaukus
    • 3 years ago
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    so \[T_A \circ T_b = T_A ( T_b(x) )\]

  11. KonradZuse
    • 3 years ago
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    mhm

  12. KonradZuse
    • 3 years ago
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    but there is no x here?

  13. KonradZuse
    • 3 years ago
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    nor does it say what kind of transformation we are performing...?

  14. UnkleRhaukus
    • 3 years ago
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    try this \[T_A\circ T_B=\text A\text B=\left[\begin{matrix}1 & -2 & 0 \\ 4 & 1 & -3 \\ 5 & 2 & 4\end{matrix}\right]\left[\begin{matrix}2 & -3 & 3 \\ 5 & 0 & 1 \\ 6 & 1 & 7\end{matrix}\right]=\]

  15. KonradZuse
    • 3 years ago
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    \[\left[\begin{matrix}-8 & -3 & 1 \\ -5 & -15 & -8 \\ 44 & -11 & 45\end{matrix}\right]\]

  16. KonradZuse
    • 3 years ago
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    wow... Is that all we needed to do? That's some BS LOL.....

  17. UnkleRhaukus
    • 3 years ago
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    looks good

  18. KonradZuse
    • 3 years ago
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    Why is that all we have to do... Why was that so easily konfusing... :'(. I hate Linear >(

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