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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]

Linear Algebra
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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
what do you mean by the little circle o
it's just what the book says, let me screen shot.

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Other answers:

what does it mean
I believe it says the composite of linear transformations.
1 Attachment
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.....
Save me :).
It gives me the answer already, but I want to know why that's the answer and how to get it...
so \[T_A \circ T_b = T_A ( T_b(x) )\]
mhm
but there is no x here?
nor does it say what kind of transformation we are performing...?
try this \[T_A\circ T_B=\text A\text B=\left[\begin{matrix}1 & -2 & 0 \\ 4 & 1 & -3 \\ 5 & 2 & 4\end{matrix}\right]\left[\begin{matrix}2 & -3 & 3 \\ 5 & 0 & 1 \\ 6 & 1 & 7\end{matrix}\right]=\]
\[\left[\begin{matrix}-8 & -3 & 1 \\ -5 & -15 & -8 \\ 44 & -11 & 45\end{matrix}\right]\]
wow... Is that all we needed to do? That's some BS LOL.....
looks good
Why is that all we have to do... Why was that so easily konfusing... :'(. I hate Linear >(

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