Question

Let T: P1---> R^2 be the function defined by the formula T(P0,P2)
a) find T (2-x)
b) find T^-1
Please, help

Answer 1

problem 17. a little bit different when in review set, it turns T(P0,P2)

Answer 2

\[T(1-2x)=(1,-1)\]

Answer 3

T x means?

Answer 4

\[p(x)=1-2x\]

Answer 5

p(x) = a0 + a1 *x

Answer 6

yup and you are given \(a_0,a_1\)

Answer 7

look at your typo!!! it is so nonsense!!
P (0) = a0
P(1) = a0 + a1
P(2) = a0 + a1*x

Answer 8

prob 17, right?

Answer 9

yes, but I am working with the same problem a little bit different from #17. That problem define T (Px)= (P0,P1) , Mine T (P(x)) = (P0,P2)

Answer 10

ok..
\[T(p(x))=(p(0),p(1))\\
{\rm when}\;p(x)=2-x\\
T(2-x)=\left(2-(0),2-(1)\right)=\mathbf{(2,1)} \]

Answer 11

yes

Answer 12

good, so, what is the question now?

Answer 13

find T^-1

Answer 14

ok, sorry, I got it.

Answer 15

\[
{\rm let}\;T^{-1}(x)=a_0+a_1x\\
p(0)=a_0\qquad p(1)=p(0)+a_1\implies a_1=p(1)-p(0)
\]

Answer 16

kid, a chain of letter, how can I know what is what?

Answer 17

I did not understadn that.

Answer 18

I don't know what's wrong with my computer, let me copy what I see in your post
T x a a x
p a p p a a p p
to me, it is no meaning

Answer 19

hold on. I think I am wrong with the inverse thing

Answer 20

It should also be another matrix.

Answer 21

ooh.. try refreshing the page

Answer 22

press the "F5" button on the keyboard

Answer 23

Ok, it makes sense now

Answer 24

I got it, kid.

Answer 25

the inverse too?

Answer 26

no, just got how to get T =(2,1)

Answer 27

ok
for inverse, we'd need the p(x)

Answer 28

and we find the standard matrix for the transformation

Answer 29

@Hoa you there?

Answer 30

yes, sir

Answer 31

so, are we taking \(p(x)=2-x\)?

Answer 32

yes

Answer 33

then\[T(p(x))=(2,1)\]
and its standard matrix is non-invertible!!

Answer 34

since the linear mapping is between two regions of different dimensions

Answer 35

is it not T^-1(2,1) = P(x)?

Answer 36

thats the only way I can think of it.

Answer 37

but, in general, you have an inverse for \(\mathbb{R}^n\rightarrow\mathbb{R}^n\)

Answer 38

I know what you mean. ok, so the conclusion is T is non-invertible matrix?

Answer 39

not "T" but the inverse matrix of the transformation does not exist.
T -> is the transformation, not the matrix -> has a system matrix "A" whose inverse may or maynot exist.
T^{-1} -> is the inverse transformation