Finding T(v) relative to B and B'

  • Thread starter Thread starter dzimitry
  • Start date Start date
  • Tags Tags
    Relative
dzimitry
Messages
4
Reaction score
0

Homework Statement



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) }


Homework Equations





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
 
Physics news on Phys.org
that's a matrix A btw, everything that was indented got shifted.
 
dzimitry said:

Homework Statement



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) }


Homework Equations





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
\begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}
will be the first column of the matrix.

= (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
 
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top