matrix of linear transformation The linear transformation $A:\mathbb{R}^2\to \mathbb{R}^2$  is given by the images of basis vectors:
$A((1,1))=(2,1)$ and $A((1,0))=(0,3)$.


*

*Find a matrix of linear transformation $A$ in the basis $(1,1), (1,0)$.

*Find $A((3,2))$.

*Find vector $x=(x_1,x_2)$ such that the matrix $\begin{pmatrix}-6 &-6\\ 3 &4\end{pmatrix}$
        is matrix of the linear transformation $A$ in the basis $x$, $(0,3)$.
Please help me about this. 
 A: If $\cal B=\{ {\bf v},{\bf w}\}$ is an ordered basis of $\Bbb R^2$, then the matrix representation, $M$, of $A:\Bbb R^2\rightarrow\Bbb R^2$ with respect to this basis (I assume you want both the domain and range to have basis $\cal B$) is the $2\times 2$ matrix that has as its first column  the coordinates of $A{\bf v}$ with respect to $\cal B$ and as its second column the coordinates of $A{\bf w}$ with respect to $\cal B$. 
What does this mean? Well, if you write a vector ${\bf x}$ in terms of this basis
$${\bf x}= c_1{\bf v}+c_2{\bf w},$$ then, setting, $[{\bf x}]_{\cal B}=[{c_1\atop c_2}]$
$$
\tag {1}[A {\bf x } ]_{\cal B} = M[{\bf x }]_{\cal B}.
$$  
That is, for $\bf x$ written in the standard basis, the coordinates of $A{\bf x}$ with respect to $\cal B$ are given by the product of the matrix $M$ with the coordinate matrix of
$\bf x$ with respect to $\cal B$.

For part 1.: 
The matrix representation of $A$ is easily found, since you were told what $A\bigl((1,1)\bigr)$ and $A\bigl((1,0)\bigr)$ were. 
We need to write $(2,1)$ and $(0,3)$ in terms of the basis $\cal B=\{(1,1),(1,0)\}$.
$$(2,1)= 1 (1,1)+1(1,0)$$
and
$$
(0,3)=  3(1,1)-3(1,0)
$$
The matrix $M$ is 
$$M=\Bigl[\, \underbrace{1\atop 1}_{ [A(1,1)]_{\cal B} } \ 
\underbrace{3\atop  -3}_{ [A(1,0)]_{\cal B} }\,\Bigr].$$

For part 2.: 
You need to write $(3,2)$ in terms of $\cal B$:
$$
(3,2)=2(1,1)+1(1,0).
$$ 
Using the matrix representation of $A$, 
$$[A\bigl((3,2 )\bigr)]_{\cal B}=\Bigl [\, {1\atop 1}\ {3\atop  -3}\,\Bigr ]\Bigl[ {2\atop 1}\Bigr]=\Bigl[{5\atop -1} \Bigr]. $$
This gives the coordinates of $A((3,2))$ with respect to $\cal B$, so
$$
A((3,2))= 5(1,1)+(-1)(1,0)=(4,5).
$$

For part 3.: 
Let $\cal B'=\{(x_1,x_2), (0,3)\}$
You know the matrix $$W= \Bigl[\,{-6\atop 3}\ {-6\atop4}\,\Bigr ] $$ is the matrix representation of $A$ with respect to $\cal B'$.
The second column of $W$ is $[ A\bigl((0,3)\bigr)]_{\cal B'}$. 
So, $$\tag{2}A((0,3))=-6(x_1,x_2)+4(0,3)=(-6x_1, -6x_2+12).$$
But you can compute $A\bigl((0,3)\bigr)$ using the matrix representation from part 1. 
We find  $[A\bigl((0,3)\bigr)]_{\cal B}$ first. Towards this end, we write $(0,3)$ in terms of the basis $\cal B$ first. Solve:
$$
(0,3) = c_1(1,1)+c_2(1,0)
$$
to obtain
$$
\eqalign{
c_1&=3\cr c_2&=-3.
}
$$
Then:
$$
[A\bigl((0,3)\bigr)]_{\cal B}=\Bigl [\, {1\atop 1}\ {3\atop  -3}\,\Bigr ]\Bigl[ {3\atop -3}\Bigr]=\Bigl[{  -6\atop  12}\Bigr].
$$
So, the coordinates of $A((0,3))$ with respect to $\cal B$ are $(-6,12)$. So,
$$
\tag{3}A((0,3))=-6(1,1)+12(1,0)= (6,  -6)
$$
Comparing equations (2) and (3) gives
$$
\eqalign{
6&=-6x_1\cr
 -6&=-6x_2+12
}
$$
This gives $x_1=-1$ and $x_2=3$.
