Question

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]

Answer 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

Answer 2

what do you mean by the little circle
o

Answer 3

it's just what the book says, let me screen shot.

Answer 4

what does it mean

Answer 5

I believe it says the composite of linear transformations.

Answer 6

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.....

Answer 7

Save me :).

Answer 8

It gives me the answer already, but I want to know why that's the answer and how to get it...

Answer 9

so
\[T_A \circ T_b = T_A ( T_b(x) )\]

Answer 10

mhm

Answer 11

but there is no x here?

Answer 12

nor does it say what kind of transformation we are performing...?

Answer 13

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

Answer 14

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

Answer 15

wow... Is that all we needed to do? That's some BS LOL.....

Answer 16

looks good

Answer 17

Why is that all we have to do... Why was that so easily konfusing... :'(. I hate Linear >(