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KonradZuse
Group Title
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]
 one year ago
 one year ago
KonradZuse Group Title
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]
 one year ago
 one year ago

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KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
what do you mean by the little circle o
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
it's just what the book says, let me screen shot.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
what does it mean
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
I believe it says the composite of linear transformations.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
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.....
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
Save me :).
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
It gives me the answer already, but I want to know why that's the answer and how to get it...
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
so \[T_A \circ T_b = T_A ( T_b(x) )\]
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
but there is no x here?
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
nor does it say what kind of transformation we are performing...?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
try this \[T_A\circ T_B=\text A\text B=\left[\begin{matrix}1 & 2 & 0 \\ 4 & 1 & 3 \\ 5 & 2 & 4\end{matrix}\right]\left[\begin{matrix}2 & 3 & 3 \\ 5 & 0 & 1 \\ 6 & 1 & 7\end{matrix}\right]=\]
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
\[\left[\begin{matrix}8 & 3 & 1 \\ 5 & 15 & 8 \\ 44 & 11 & 45\end{matrix}\right]\]
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
wow... Is that all we needed to do? That's some BS LOL.....
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
looks good
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.1
Why is that all we have to do... Why was that so easily konfusing... :'(. I hate Linear >(
 one year ago
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