Standard matrix A of T? Help please. What would be the standard matrix of A? I know how to do number 2 and 3 but I'm just having trouble with A. I asked this earlier but I lost my account and I'm not sure if I posted correctly. I am sorry.

 A: The matrix of $T$ is $A = \Big(\begin{array} &a & b \\ c &d \end{array} \Big)$.
What you are given is $\Big(\begin{array} &a & b \\ c &d \end{array} \Big) \Big(\begin{array} &1 \\ 3 \end{array} \Big) = \Big(\begin{array} &-2 \\ -6 \end{array} \Big)$ and $\Big(\begin{array} &a & b \\ c &d \end{array} \Big)\Big(\begin{array} &2 \\ 5 \end{array} \Big) = \Big(\begin{array} &2 \\ 5 \end{array} \Big)$. That's a system of four linear equations, that shouldn't be a problem.
A: Let $T(x)$ be the linear transformation. Then there exists a unique A such that
$T(x)=Ax$ for all x in $R^2$
In first example:
 $x = \Big(\begin{array} &1\\ 3\end{array} \Big)$.
$T(x)=-2(x)$// Where x is a column vector.
A standard matrix is the $m*n$ matrix whose jth column is the vector $T(ej)$, where ej is the jth  column of the identity matrix in $R^n$:
To find standard matrix:
$T(e1)=-2e1 =-2\Big(\begin{array} &1 \\ 0 \end{array} \Big)=\Big(\begin{array} &-2 \\ 0 \end{array} \Big)$
$T(e2)=-2e2 =-2\Big(\begin{array} &0 \\ 1 \end{array} \Big)=\Big(\begin{array} &0 \\ -2 \end{array} \Big)$
$A=\Big(\begin{array} &T(e1)...&T(en) \\  \end{array} \Big)$
$A=\Big(\begin{array} &-2 &0\\0 &-2 \end{array} \Big)$, A is the 2*2 matrix.
This is the standard matrix.
Similarly you can solve for second part.
