<s>[INST] Finding the matrix associated with a linear map 
Find the matrix associated with the linear map $f:R^2 \rightarrow R^2$ defined by $f(x,y)=(3x-y,y-x)$ with respect to the ordered basis ${(1,0),(1,1)}$

Let the matrix be $A$ and let $f(x)=AX$ where 
$$X=\begin{bmatrix} 
x \\
y 
\end{bmatrix}$$
and 
$$A=\begin{bmatrix} 
a & b \\
c & d 
\end{bmatrix}$$
I tried solving for $a,b,c,d$ by using the basis vectors as $(x,y)$. That is, I took $(x,y)=(1,0)$ and $(x,y)=(1,1)$ to find $a,b,c,d$. But I am not getting the given answer.
 [/INST] Regarding to the standard basis $e_i$ the map has the matrix
$$
A = 
\begin{pmatrix}
3 & -1 \\
-1 & 1
\end{pmatrix}
$$
because 
$$
f(x) 
= x A
= (x_1 e_1 + x_2 e_2) A
= x_1 e_1 A + x_2 e_2 A
= (x_1, x_2) 
\begin{pmatrix}
e_1 A \\
e_2 A
\end{pmatrix}
\\
e_1 = (1,0):\quad f(1,0) = (3,-1) = e_1 A = (1,0) A = (a_{11} a_{12})\\
e_2 = (0,1):\quad f(0,1) = (-1,1) = e_2 A = (0,1) A = (a_{21} a_{22})
$$
What you are looking for is
$$
B = M A M^{-1}
$$
$M$ first transforms a vector regarding the basis $b_1 = (1,0)$ and $b_2 = (1,1)$ into standard coordinates (basis $e_i$). Then it applies the function $f$ via the matrix $A$ (representing $f$ if you use it with vectors regarding standard coordinates) and finally transforms the result into the same vector regarding the basis $b_i$ via $M^{-1}$. 
The combined matrices have the same effect as $f$ for vectors regarding the basis $b_i$.
As the $b_i$ regarding themselves have the coordinates $(1,0)$ and $(0,1)$ these vectors must be mapped by $M$ to $(1,0)$ and $(1,1)$, thus
$$
M = 
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}
\quad
M^{-1} = 
\begin{pmatrix}
1 & 0 \\
-1 & 1
\end{pmatrix}
$$
and we calculate
\begin{align}
B
&= M A M^{-1} \\
&=
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}
\begin{pmatrix}
3 & -1 \\
-1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
-1 & 1
\end{pmatrix}
\\
&=
\begin{pmatrix}
3 & -1 \\
2 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
-1 & 1
\end{pmatrix}
\\
&=
\begin{pmatrix}
4 & -1 \\
2 & 0
\end{pmatrix}
\end{align}
Test:
$e_1 = (1,0)$ regarding basis $b_i$ still has coordinates $(1,0)$,
as $(1,0) = 1 \cdot b_1 + 0 \cdot b_2$.
Then
$$
e_1 B = 
(1,0)
\begin{pmatrix}
4 & -1 \\
2 & 0
\end{pmatrix}
= (4,-1)
$$
And indeed 
$$
4 b_1 - b_2 = 4 (1,0) - (1,1) = (4,0)-(1,1) = (3,-1)
$$
Further $e_2 = (0,1)$ regarding standard coordinates has coordinates $(-1,1)$ regarding the $b_i$:
$$
-1 \cdot b_1 + 1 \cdot b_2 = (-1,0) + (1,1) = (0,1)
$$
This is mapped as:
$$
(-1,1)
\begin{pmatrix}
4 & -1 \\
2 & 0
\end{pmatrix}
= (-2, 1)
$$
Calculating back to standard coordinates:
$$
-2 b_1 + b_2 = -2 (1,0) + (1,1) = (-2,0) + (1,1) = (-1,1)
$$
which is where $e_2$ should mapped to by $f$.
Note:
The above assumed from your question that you have to use row vectors. If you use column vectors you
have to transpose the matrices and vectors:
$$
f(x) = A x 
$$
as $A = A^T$ is symmetric. Further
$$
B = (MAM^{-1})^T 
= (M^{-1})^T A^T M^T
= (M^T)^{-1} A M^T
=
\begin{pmatrix}
4 & 2 \\
-1 & 0
\end{pmatrix}
$$
</s>