I am just not certain about the technique you used (I think it's partially do to the notation we have to use on these boards).
Specifically I am not sure how you formed T_A or T_1 (why is T_A = (5x_1 + 4x_2, 3x_1 + 6x_2, ?, ?)).
If you could elaborate on this process a little further, I'd appreciate it.
Paul Sperry wrote: > In article <email@example.com>, > <"firstname.lastname@example.org"> wrote: > > > Let T1: R4 --> R2 and T2: R2 --> R4. > > > > If T1(x1, x2, x3, x4) = (x1, x2) and T2(x1, x2) = (x1, x2, 0, 0) and > > Ta: R4 --> R4 be defined by T(x) Ax, where > > > > A is 4x4 with the following rows from top to bottom > > [5 4 -1 9] > > [3 6 1 5] > > [5 2 -1 4] > > [1 -2 -3 5] > > > > Find T1 o TA o T2. > > > > Thanks. > > > > Adjusting your notation a little: ' means transpose. > > T_2((x_1, x_2)') = (x_1, x_2, 0, 0)'. > > T_A((x_1, x_2, 0, 0)') = A(x_1, x_2, 0, 0)' = > (5x_1 + 4x_2, 3x_1 + 6x_2, ?, ?)'. > > T_1((5x_1 + 4x_2, 3x_1 + 6x_2, ?, ?)') = (5x_1 + 4x_2, 3x_1 + 6x_2)' > > If you would ask the questions from the "Urgent help" thread the way > you did this one, you would have a lot better luck. I, too, am among > those who are not willing to download a pdf file; I'm not offended, I > just won't do it. > > -- > Paul Sperry > Columbia, SC (USA)