Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
$A=\begin{pmatrix} -3 & -3 \\ -2 & 4 \end{pmatrix}  $
 A: Denote
$$A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}  $$
and solve the system of $4$ equations
$$A\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $$
$$A\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$$
for the unknowns $a,b,c$ and $d$.
A: The matrix A say it has entries $a_{1,1}, a_{1,2}, a_{2,1}, a_{2,2} $. You know this is the correct size by definition of matrix multiplication and the size of the vectors given (input and output).
To solve it simply multiply this matrix by the two vectors $ (3, -2)$ and $ ( 3,0)$ and put this equal to the output of the linear map. You'll get a system of equations and then you simply solve it and get the answer.
Remember that you can do that you can use the matrix representing the linear transformation and apply it to the vectors by simple matrix multiplication.
A: One possibility is "brute force" solution: you know $A\pmatrix{3 \\ -2} = \pmatrix{-3 \\ -14}, A\pmatrix{3 \\ 0} = \pmatrix{-9 \\ -6}$, that's four equations for four unknown elements, solvable in a number of ways. This approach is inelegant and quickly becomes hard for bigger dimensionality, but is perfectly doable.
The other is that you know matrix of $\phi$ in basis $b' = (\pmatrix{3 \\ -2}, \pmatrix{3 \\ 0})$ (it is $A' = \pmatrix{-3 & -9 \\ -14 & -6}$) and you want to learn what matrix it has in basis $b_0 = (\pmatrix{1 \\ 0}, \pmatrix{0 \\ 1})$. That would be $A'*\pmatrix{3 & 3 \\ -2 & 0}^{-1} = A'*\pmatrix{0 & -{1 \over 2} \\ 1 \over 3 & 1 \over 2} = \pmatrix{-3 & -3 \\ -2 & 4}$.
A: We note that when we are constructing a matrix for a linear map, we need to know the matrix related to which basis. Here, if you consider the vectors 
$\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ 
as the basis of your vector space, the given matrix by you is not the correct one. Actually, as we have
$$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix}=7\times \begin{pmatrix} 3 \\ -2 \end{pmatrix}-8\times \begin{pmatrix} 3 \\ 0 \end{pmatrix}$$
$$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}=3\times \begin{pmatrix} 3 \\ -2 \end{pmatrix}-6\times \begin{pmatrix} 3 \\ 0 \end{pmatrix}$$
The matrix should be (according to this basis!)
$$A=\begin{pmatrix} 7 & 3 \\ -8 & -6 \end{pmatrix}$$
Perhaps your given matrix is related to another basis.
A: Let's go through the fundamental ideas here from the beginning:
If you multiply a matrix $$A=\pmatrix{a_{11} & a_{12} & \cdots & a_{1m} \\ a_{21} & a_{22} & \cdots & a_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nm}}$$ by the vector $e_1 = \pmatrix{1 \\ 0 \\ \vdots \\ 0}$, what do you get?  Let's see:
$$Ae_1 = \pmatrix{a_{11} & a_{12} & \cdots & a_{1m} \\ a_{21} & a_{22} & \cdots & a_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nm}}\pmatrix{1 \\ 0 \\ \vdots \\ 0} = \pmatrix{a_{11} + 0a_{12} + \cdots + 0a_{1m} \\ a_{21} + 0a_{22} + \cdots + 0a_{2m} \\ \vdots \\ a_{n1} + 0a_{n2} + \cdots + 0a_{nm}} = \pmatrix{a_{11} \\ a_{21} \\ \vdots \\ a_{n1}}$$
Notice anything about this vector?  It's just the first column of $A$.  That's interesting.
I claim that if you multiply $A$ by $e_2 = \pmatrix{0 \\ 1 \\ 0 \\ \vdots \\ 0}$, you'd get the second column of $A$.  And if you multiplied $A$ by $e_i$, which is a column vector with zeroes everywhere except for a 1 in the $i$th row, then you'd get the $i$th column of $A$.  Try it out and see what you get.
So, with that in mind, we can see that any matrix $A$ can be thought of as  $$\pmatrix{Ae_1 & Ae_2 & \cdots & Ae_m}$$
That may not seem very useful but let's consider your function $\phi$.  If we could figure out what $\phi(e_1)$ and $\phi(e_2)$ are then the matrix representation of $\phi$ would just be $$\pmatrix{\phi(e_1) & \phi(e_2)}$$
So let's see if we can do that.  We know that $\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix}$ and $\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$ and we know that $\phi$ is linear.  So let's use that linearity.  The second equation almost looks like $\phi\pmatrix{1 \\ 0}$, so let's start with that one:
$$\phi\pmatrix{3 \\ 0} = \phi\left[3\pmatrix{1 \\ 0}\right] = 3\phi\pmatrix{1 \\ 0} = \pmatrix{-9 \\ -6} \\ \implies \phi\pmatrix{1 \\ 0} = \frac 13\pmatrix{-9 \\ -6} = \pmatrix{-3 \\ -2}$$
Using that, let's figure out $\phi\pmatrix{0 \\ 1}$:
$$\phi\pmatrix{3 \\ -2} = \phi\left[3\pmatrix{1 \\ 0} -2\pmatrix{0 \\ 1}\right] = 3\phi\pmatrix{1 \\ 0}-2\phi\pmatrix{0 \\ 1} = \pmatrix{-9 \\ -6}-2\phi\pmatrix{0 \\ 1} = \pmatrix{-3 \\ -14} \\ \implies \phi\pmatrix{0 \\ 1} = -\frac 12\left[\pmatrix{-3 \\ -14}-\pmatrix{-9 \\ -6}\right] = \pmatrix{-3 \\ 4}$$
So then we can see that $$\bbox[5px,border:2px solid red]{A=\pmatrix{-3 & -3 \\ -2 & 4}}$$
