This is a problem from 2012 mid-term exam. And this is also a practice for my upcoming mid-term in 1 week.
The problem goes like this:
if $T:R^2\rightarrow R^2$ is a linear transformation so that.
$$T(\begin{pmatrix}3 \\ 0\end{pmatrix})=\begin{pmatrix}9 \\ 3\end{pmatrix},T(\begin{pmatrix}0 \\ -\frac{1}{2}\end{pmatrix})=\begin{pmatrix}1 \\ -1\end{pmatrix}$$
What is the matrix of $T$
why can't I solve it like this?:
$$e_1=\begin{pmatrix}1 \\0\end{pmatrix}, e_2=\begin{pmatrix}0 \\ 1\end{pmatrix}$$
$$\begin{pmatrix}3 & 0&9 & 1\\ 0 &-\frac{1}{2} &3 &-1\end{pmatrix}$$
rref
$$\begin{pmatrix}1 & 0&3 & \frac{1}{3}\\ 0 &1 &-6 &2\end{pmatrix}$$
Since if I manage to obtain the identity matrix on the left, shouldn't it be sufficient enought to obtain the matrix $T$?
which in my case is $$T=\begin{pmatrix}3&\frac{1}{3} \\ -6 & 2\end{pmatrix}$$
but according to the exam solutions this is wrong and should be
$$\begin{pmatrix}3 & -2 \\ 1 &2\end{pmatrix}$$
If you want to row reduce, use $$\begin{pmatrix} 3 & 0 & T(3,0) \\ 0 & -1/2 & T(0,-1/2) \end{pmatrix}$$ Then your first step, dividing by 3, you calculate $(1/3)T(3,0) = (3,1)=T(1,0)$. So row reducing you end up with $$\begin{pmatrix} 1 & 0 & T(e_1) \\ 0 & 1 & T(e_2) \end{pmatrix}$$ And $T(e_1), T(e_2)$ are the columns of matrix $T$. Although it may be easier to just find $T(e_1)$, $T(e_2)$ using linearity as opposed to thinking about row reduction.