Find $2\times 2$ matrix of linear transformation $T$ Given:
\[
T\left(\begin{bmatrix}-1 \\ -2\end{bmatrix}\right) = \begin{bmatrix}17 \\ 11\end{bmatrix} \text{  and  } T\left(\begin{bmatrix}2 \\ 3\end{bmatrix}\right) = \begin{bmatrix}-30 \\ -17\end{bmatrix}
\]
Find a matrix such that:
\[ T(v) = \begin{bmatrix}? & ? \\ ? & ? \end{bmatrix}\cdot v \]
I have no idea where to start... Any help will be much appreciate it.
 A: Hint: $B=\{\begin{pmatrix}-1\\
-2
\end{pmatrix},\begin{pmatrix}2\\
3
\end{pmatrix}\}$ is a basis for $\mathbb{R}^{2}$. Can you write $[T]_{B}$ ? 
Can you change basis from $B$ to $B'=\{\begin{pmatrix}1\\
0
\end{pmatrix},\begin{pmatrix}0\\
1
\end{pmatrix}\}$ to get $[T]_{B'}$which is the desired result ?
A: The answer is $\begin{bmatrix} T(e_1) & T(e_2) \end{bmatrix}$ where $e_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$.  We just need to figure out what $T(e_1)$ and $T(e_2)$ are.
Let $v_1 = \begin{bmatrix} -1 \\ -2 \end{bmatrix}$ and $v_2 = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$.  Notice that
\begin{equation}
e_1 = 3 v_1 + 2 v_2
\end{equation}
and
\begin{equation}
e_2 = -2 v_1 - v_2.
\end{equation}
These equations allow us to compute $T(e_1)$ and $T(e_2)$.
A: Let's write $$\begin{bmatrix} a & b \\ c & d\end{bmatrix}$$ for the matrix we want to find. Then the two given equations read 
\begin{align*}
  -a - 2b &= 17\\
  -c - 2d &= 11\\[3mm]
  2a + 3b &= -30\\
  2c + 3d &= -17
\end{align*}
We have to solve this system, so let's first look at the equations for $a$ and $b$
\begin{align*}
  -a - 2b &= 17\\
  2a + 3b &= -30
\end{align*}
Adding the first two times to the second gives
\begin{align*}
  -a - 2b &= 17\\
       -b &= 4
\end{align*}
Hence $b = -4$ and $a = -9$. The second system reads
\begin{align*}
  -c - 2d &= 11\\
  2c + 3d &= -17
\end{align*}
Adding again, we have
\begin{align*}
  -c - 2d &= 11\\
       -d &= 5
\end{align*}
Hence $d = -5$ and $c = -1$.
So $$ \begin{bmatrix} -9 & -4 \\ -1 & -5\end{bmatrix} $$ is the matrix we looked for.
A: The matrix $A=\begin{pmatrix}17&-30\\11&-17\end{pmatrix}$ defines your linear map $T$, but using the basis $B=[b_1,b_2]=[\binom{-1}{-2},\binom23]$ (rather than the canonical basis $E=[e_1,e_2]=\binom10,\binom01]$) at departure, while using the canonical basis $E$ at arrival: this means that $$T(xb_1+yb_2)=A\cdot\binom xy$$ (you can check that this is what is given, for $(x,y)=(1,0)$ and for $(x,y)=(0,1)$; the general case follows by linearity). I would personally write this  (but the notation is not widely used) as $A={}_C\!\operatorname{Mat}_B(T)$; the idea of this notation is that the basis at departure is written at the right where the vector argument "enters", and the basis at arrival is written on the left where the result "leaves" the matrix multiplication. You want to know $M={}_C\!\operatorname{Mat}_C(T)$, the matrix when using the canonical bases all the time.
This means you want to take a vector $\binom xy$ expressed in $C$, first transform it into coordinates on $B$ (find $x',y'$ such that $\binom xy=x'b_1+y'b_2$) and then compute $A\cdot\binom{x'}{y'}$. The opposite coordinate transformation is easy: given $x',y'$ one has $\binom xy=x'b_1+y'b_2=\begin{pmatrix}-1&2\\-2&3\end{pmatrix}\cdot\binom{x'}{y'}$. But then we must have $\binom{x'}{y'}=\begin{pmatrix}-1&2\\-2&3\end{pmatrix}^{-1}\cdot\binom xy$, and one can compute
$$
  \begin{pmatrix}-1&2\\-2&3\end{pmatrix}^{-1} =
  \begin{pmatrix}3&-2\\2&-1\end{pmatrix}.
$$
Now the compound operation of first doing the coordinate transform from $\binom xy$ to $\binom{x'}{y'}$ and then applying the linear map $T$ through its matrix $A$ has as matrix $M$ the matrix product
$$
  M = A \cdot\begin{pmatrix}-1&2\\-2&3\end{pmatrix}^{-1}
  =\begin{pmatrix}17&-30\\11&-17\end{pmatrix}\cdot
   \begin{pmatrix}3&-2\\2&-1\end{pmatrix}
  =\begin{pmatrix}-9&-4\\-1&-5\end{pmatrix}.
$$
