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Math
Emeritus
Thanks
PF Gold
P: 38,706
 Quote by dzimitry 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
$$\begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}$$
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 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution