A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 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]
 2 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]

This Question is Closed

KonradZuse
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1what do you mean by the little circle o

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

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

KonradZuse
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1so \[T_A \circ T_b = T_A ( T_b(x) )\]

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

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

UnkleRhaukus
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1wow... Is that all we needed to do? That's some BS LOL.....

KonradZuse
 2 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 >(
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.