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KonradZuse
 4 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]
KonradZuse
 4 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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KonradZuse
 4 years ago
Best ResponseYou've already chosen the best response.1A =\[\left[\begin{matrix}1 & 2 & 0 \\ 4 & 1 & 3 \\ 5 & 2 & 4\end{matrix}\right]\] B = \[\left[\begin{matrix}2 & 3 & 3 \\ 5 & 0 & 1 \\ 6 & 1 & 7\end{matrix}\right]\] @satellite73

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1what do you mean by the little circle o

KonradZuse
 4 years ago
Best ResponseYou've already chosen the best response.1it's just what the book says, let me screen shot.

KonradZuse
 4 years ago
Best ResponseYou've already chosen the best response.1I believe it says the composite of linear transformations.

KonradZuse
 4 years ago
Best ResponseYou've already chosen the best response.1http://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.....

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

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1so \[T_A \circ T_b = T_A ( T_b(x) )\]

KonradZuse
 4 years ago
Best ResponseYou've already chosen the best response.1but there is no x here?

KonradZuse
 4 years ago
Best ResponseYou've already chosen the best response.1nor does it say what kind of transformation we are performing...?

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

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

KonradZuse
 4 years ago
Best ResponseYou've already chosen the best response.1wow... Is that all we needed to do? That's some BS LOL.....

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