<s>[INST] 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}  $
 [/INST] 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 $$\bbox[5px,border:2px solid red]{A=\pmatrix{-3 & -3 \\ -2 & 4}}$$
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