# Finding T(v) relative to B and B'

1. ### dzimitry

4
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.

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

2. Relevant equations

3. The attempt at a solution

2. ### dzimitry

4
that's a matrix A btw, everything that was indented got shifted.

3. ### HallsofIvy

40,809
Staff Emeritus
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.

4. ### dzimitry

4
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?