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-   -   Finding T(v) relative to B and B' (http://www.physicsforums.com/showthread.php?t=421161)

dzimitry Aug9-10 03:00 AM

finding T(v) relative to B and B'
 
1. The problem statement, all variables and given/known data

find T(v) using the matrix relative to B and B'

T(x, y, z) = (2x, x + y, y + z, x + z)
v = (1, -5, 2)
B = { (2, 0, 1), (0, 2, 1), (1, 2, 1) }
B' = { (1, 0, 0, 1), (0, 1, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0) }


2. Relevant equations



3. The attempt at a solution

T(2, 0, 1) = (4, 2, 1, 3)
= 4(1, 0, 0, 1) + 2(0, 1, 0, 1) + 1(1, 0, 1, 0) + 3(1, 1, 0, 0)
= (8, 5, 1, 6)
T(0, 2, 1) = (0, 2, 3, 1)
= (4, 3, 3, 2)
T(1, 2, 1) = (2, 3, 3, 2)
= (7, 5, 3, 5)

A = 8 4 7
5 3 5
1 3 3
6 2 5

Av = (2, 0, -8, 6)

but if the person I am checking against is right, the answer should be (2, -4, -3, 3)

I am confused as to if I can even use the method I am using in this case.

Thanks in advance
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

dzimitry Aug9-10 03:01 AM

Re: finding T(v) relative to B and B'
 
that's a matrix A btw, everything that was indented got shifted.

HallsofIvy Aug9-10 08:13 AM

Re: finding T(v) relative to B and B'
 
Quote:

Quote by dzimitry (Post 2832607)
1. The problem statement, all variables and given/known data

find T(v) using the matrix relative to B and B'

T(x, y, z) = (2x, x + y, y + z, x + z)
v = (1, -5, 2)
B = { (2, 0, 1), (0, 2, 1), (1, 2, 1) }
B' = { (1, 0, 0, 1), (0, 1, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0) }


2. Relevant equations



3. The attempt at a solution

T(2, 0, 1) = (4, 2, 1, 3)
= 4(1, 0, 0, 1) + 2(0, 1, 0, 1) + 1(1, 0, 1, 0) + 3(1, 1, 0, 0)

No, (4, 2, 1, 3) is NOT equal to (8, 5, 1, 6)! You are doing this backwards. You want to find numbers, a, b, c, d, such that (4, 2, 1, 3)= a(1, 0, 0, 1)+ b(0, 1, 0, 1)+ c(1 , 0, 1, 0)+ d(1, 1, 0, 0). That is you jeed to solve a+ c+ d= 4, b+ d= 2, c= 1, and a+ b= 3.
Then
[tex]\begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}[/tex]
will be the first column of the matrix.

Quote:

= (8, 5, 1, 6)
T(0, 2, 1) = (0, 2, 3, 1)
= (4, 3, 3, 2)
T(1, 2, 1) = (2, 3, 3, 2)
= (7, 5, 3, 5)

A = 8 4 7
5 3 5
1 3 3
6 2 5

Av = (2, 0, -8, 6)

but if the person I am checking against is right, the answer should be (2, -4, -3, 3)

I am confused as to if I can even use the method I am using in this case.

Thanks in advance
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

dzimitry Aug9-10 11:28 AM

Re: finding T(v) relative to B and B'
 
ok that makes sense...and for the vector v = (1, -5, 2), do I need to solve a system like
(1, -5, 2) = a(2, 0, 1) + b(0, 2, 1) + c(1, 2, 1) and use (a, b, c) as my v and multiply that by A?


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