##### Carl's Answer to PM #7 33A
CarlHarmon
It seems like there are a variety of answers for #7. Here is my answer with the work. Hopefully we can reach a consensus. Basically, this problem can be broken into three steps: 1) Find T for each \beta 2) Get the arbitrary values in terms of \beta 3) Factor and substitute First, [T(1,1)]_\gamma = $\left( \begin{array}{ccc} 1 \\ 3 \end{array} \right)$ T(1,1) = $\left( \begin{array}{ccc} 2 \\ 4 \end{array} \right)$ [T(1,-1)]_\gamma = $\left( \begin{array}{ccc} 1 \\ 2 \end{array} \right)$ T(1,1) = $\left( \begin{array}{ccc} 3 \\ 2 \end{array} \right)$ Now, solve for a and b in terms of \beta: $\left( \begin{array}{ccc} 1 & 1 & | a\\ 1 & -1 & |b \end{array} \right)$ = $\left( \begin{array}{ccc} 1 & 0 & | (a+b)/2\\ 0 & -1 & |(a-b)/2 \end{array} \right)$ Finally, T(a,b) = T((a+b)/2 $\left( \begin{array}{ccc} 1 \\ 1 \end{array} \right)$ + (a-b)/2 $\left( \begin{array}{ccc} 1 \\ -1 \end{array} \right)$ = (a+b)/2 * T$\left( \begin{array}{ccc} 1 \\ 1 \end{array} \right)$ + (a-b)/2 * T$\left( \begin{array}{ccc} 1 \\ -1 \end{array} \right)$ = (a+b)/2 * $\left( \begin{array}{ccc} 2 \\ 4 \end{array} \right)$ + (a-b)/2 * $\left( \begin{array}{ccc} 3 \\ 2 \end{array} \right)$ = $\left( \begin{array}{ccc} (5a-b)/2 \\ 3a-b \end{array} \right)$
CarlHarmon: May 4, 2015, 6:58 p.m.
Wow, I apologize for the formatting.