Question

Consider the basis S = {v1, v2, v3} for R3, where v1 = (1, 2, 1), v2 = (2, 9, 0), and v3 = (3, 3, 4), and let T:R3 → R2 be the linear operator such that T(v1) = (1, 0), T(v2) = (− 1, 1), T(v3) = (0, 1) Find a formula for T(x1, x2, x3), and use that formula to find T(9, 80, 9).

Answer 1

so we have a transformation from a 3 vector to a 2 vector

Answer 2

ok so far we have

Answer 3

Consider the basis S = {v1, v2, v3} for R3, where v1 = (1, 2, 1), v2 = (2, 9, 0), and v3 = (3, 3, 4), and let T:R3 -> R2 be the linear operator such that T(v1) = (1, 0), T(v2) = (- 1, 1), T(v3) = (0, 1) Find a formula for T(x1, x2, x3), and use that formula to find T(9, 80, 9).

Note i will use capital letters for bold letters
We first express X = (x1,x2,x3) as a linear combination of V1 = (1,2,1) , V2 = (2,9,0) and V3 = (3,3,4).

If we write

(x1,x2,x3) = k1 ( 1,2,1) + k2( 2,9 ,0) + k3 (3,3,4)

we obtain

1*k1 + 2*k2 + 3*k3 = x1
2*k1 + 9*k2 + 4*k3 = x2
1*k1+ 0*k2 + 4*k3 = x3

Answer 4

now we do guassian elimination

Answer 5

by the way, this is not the easiest way to solve this . this is the intro to the chapter im reading

Answer 6

ok so let me get my maple computer

Answer 7

err software,

Answer 8

you still with me?

Answer 9

check over for any errors

Answer 10

Yes. yes. I'm following..

Answer 11

and i already made a mistake :)

Answer 12

should be

1*k1 + 2*k2 + 3*k3 = x1
2*k1 + 9*k2 + 3*k3 = x2
1*k1+ 0*k2 + 4*k3 = x3

Answer 13

I kinda did something similar to your solution and i got (505, -124) but i'm not sure i did it right.. I happen to get the values of k1,k2, and k3 in terms of x1,x2, and x3..

Answer 14

the reduced matrix is
[[1,0,0, 21 x3-36 x1+8 x2],
[0,1,0, -3 x3+5 x1-x2 ]
[0,0,1, -5 x3+9 x1-2 x2]]

Answer 15

k1 = 21x3 - 36x1 + 8x2
k2 =- 3 x3+5 x1-x2
k3 = -5 x3+9 x1-2 x2

Answer 16

How did you get the reduced matrix? I got lost

Answer 17

I used maple software.
don't worry, we can go back to this
lets see if this solution works


(x1,x2,x3) = k1* ( 1,2,1) + k2*( 2,9 ,0) + k3*(3,3,4)

substitute

Answer 18

currently using maple 18 , thats what its called

Answer 19

We found that

k1 = 21x3 - 36x1 + 8x2
k2 =- 3 x3+5 x1-x2
k3 = -5 x3+9 x1-2 x2

We had before
(x1,x2,x3) = k1* ( 1,2,1) + k2*( 2,9 ,0) + k3*(3,3,4)

substitute

(x1,x2,x3) = (21x3 - 36x1 + 8x2)* ( 1,2,1) +( - 3 x3+5 x1-x2 ) *( 2,9 ,0) + (-5 x3+9 x1-2 x2) *(3,3,4)

Answer 20

T(x1,x2,x3) = (21x3 - 36x1 + 8x2) T(v1) + ( - 3 x3+5 x1-x2 )* T(v2) + (-5 x3+9 x1-2 x2) *T(v3)

Answer 21

T(v1) = (1, 0), T(v2) =(1, 1), T(v3) = (0, 1)

T(x1,x2,x3) = (21x3 - 36x1 + 8x2) *(1,0) + ( - 3 x3+5 x1-x2 )* (-1,1) + (-5 x3+9 x1-2 x2) *(0,1)

Answer 22

T(x1,x2,x3)
= ( (21x3 - 36x1 + 8x2) *(1)+ ( - 3 x3+5 x1-x2 )* (-1) + (-5 x3+9 x1-2 x2) *(0) , (21x3 - 36x1 + 8x2) *(0) + ( - 3 x3+5 x1-x2 )* (1) + (-5 x3+9 x1-2 x2) *(1))

Answer 23

that simplifies to

Answer 24

T(x1,x2,x3)
= ( (21x3 - 36x1 + 8x2) *(1)+ ( - 3 x3+5 x1-x2 )* (-1) , ( - 3 x3+5 x1-x2 )* (1) + (-5 x3+9 x1-2 x2) *(1) )

Answer 25

this simplifies to

Answer 26

T(x1,x2,x3)
= ( 21x3 - 36x1 + 8x2 + 3 x3-5 x1+x2 , - 3 x3+5 x1-x2 -5 x3+9 x1-2 x2 )

Answer 27

505,-124?

Answer 28

T(x1,x2,x3)
= ( 24x3 - 41x1 + 9x2 , -8 x3+14 x1-3x2 )

Answer 29

( - 3 x3+5 x1-x2 )* (-1) , ( - 3 x3+5 x1-x2 )* (1) <-- So this will not be cancelled. </3

Answer 30

woops
V1 = (1,2,1)

Answer 31

lets test this show that
T(V1) = (1,0)

ie show that T ( 1,2,1) = (1,0)

Answer 32

yes it works, just checked :)

Answer 33

those are separated by commas, so they are in different places
like

( x , y )

Answer 34

T(9, 80, 9).

plug into
T(x1,x2,x3) = ( 24x3 - 41x1 + 9x2 , -8 x3+14 x1-3x2 )

Answer 35

So it is 567, -186 :D

Answer 36

correct

Answer 37

Thank youu! <3

Answer 38

ok lets streamline this approach

Answer 39

or did you want to go back to see the gaussian elimination

Answer 40

It's okay I guess. I followed your solution so i guess understand. :D

Answer 41

1*k1 + 2*k2 + 3*k3 = x1
2*k1 + 9*k2 + 3*k3 = x2
1*k1+ 0*k2 + 4*k3 = x3


R2 ' = R2 + (-2)*R1

R3 ' = R3 + (-1)*R1

Answer 42

i will be back later :) goodbye