This by the way is called the Hill Cipher from way back in 1929 which is very trivial by today's standard. Super easy to break because of its linearity. It is only used for elementary pedagogical purposes nowadays.
Usually we need to know the encryption scheme and then we attempt to figure out the specific key used. So in this case you believe that the hill cipher is being used and you have given us some ciphertext with its known plain text but we only have three pairs and there are still too many possibilities about the specific scheme without any more information about the scheme.
1.First, a number mapping to a number under this scheme doesn't make sense at all. It must be a vector mapping to the same size vector. And it is not the 1x1 case. So I will make an educated guess here that these are probably 5-vectors being mapped to 5-vectors because all of the numbers have 5 digits in them. This then forces the key-matrix to be a 5x5 matrix like this
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] \cdot \left[\begin{array} .6\\ 2\\ 4\\ 0\\ 3 \end{array}\right] = \left[\begin{array} .7\\ 4\\ 2\\ 2\\ 2 \end{array}\right] $
2.We have no idea which number system the encryption is being done in. Usually with hill cipher (and many many others) in the English language we work mod 26 (or mod 29 because 29 is prime so it makes inversion easier by adding three more punctuation characters). Here it may be that you are working in the real numbers. But I can also interpret it as mod 26 (or mod 29).
3.Assuming that the key was 5x5, we need two more pairs and you only gave us two. The number of known ciphertext and plaintext pairs needed to break a Hill cipher depends on the matrix size which is always a square for invertibility. So for an $n\times n$ key-matrix we need $n$ pairs of vectors with each being of size $n$. Otherwise (depending on the number system) we may have infinite possibilities (like in the real case) or more than one but finitely many possibilities (like in the mod 26 or mod 29 case).
So just for the sake of answering your question I will assume that we are working in the real field and the key was 5x5. We already have three pairs and I will just make up two more pairs, let's say $10000 \rightarrow 11111$ and $11000 \rightarrow 22222$.
So now we have $\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] \cdot \left[\begin{array} .6\\ 2\\ 4\\ 0\\ 3 \end{array}\right] = \left[\begin{array} .7\\ 4\\ 2\\ 2\\ 2 \end{array}\right] $
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] \cdot \left[\begin{array} .6\\ 2\\ 4\\ 5\\ 1 \end{array}\right] = \left[\begin{array} .7\\ 4\\ 2\\ 1\\ 1 \end{array}\right] $
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] \cdot \left[\begin{array} .4\\ 4\\ 2\\ 7\\ 9 \end{array}\right] = \left[\begin{array} .7\\ 4\\ 2\\ 0\\ 8 \end{array}\right] $
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] \cdot \left[\begin{array} .1\\ 0\\ 0\\ 0\\ 0 \end{array}\right] = \left[\begin{array} .1\\ 1\\ 1\\ 1\\ 1 \end{array}\right] $
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] \cdot \left[\begin{array} .1\\ 1\\ 0\\ 0\\ 0 \end{array}\right] = \left[\begin{array} .2\\ 2\\ 2\\ 2\\ 2 \end{array}\right] $
Putting the five together, we get
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] \cdot \left[\begin{array} .6&6&4&1&1\\ 2&2&4&0&1\\ 4&4&2&0&0\\ 0&5&7&0&0\\ 3&1&9&0&0 \end{array}\right] = \left[\begin{array} .7&7&7&1&2\\ 4&4&4&1&2\\ 2&2&2&1&2\\ 2&1&0&1&2\\ 2&1&8&1&2 \end{array}\right] $
Then you just multiply the inverse of the plaintext matrix on the right on both sides and you have your key. Working in the real field, the inverse and hence the key is
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] = \left[\begin{array} .7&7&7&1&2\\ 4&4&4&1&2\\ 2&2&2&1&2\\ 2&1&0&1&2\\ 2&1&8&1&2 \end{array}\right] \cdot \left[\begin{array} .6&6&4&1&1\\ 2&2&4&0&1\\ 4&4&2&0&0\\ 0&5&7&0&0\\ 3&1&9&0&0 \end{array}\right]^{-1} $
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] = \frac{1}{206} \left[\begin{array} .7&7&7&1&2\\ 4&4&4&1&2\\ 2&2&2&1&2\\ 2&1&0&1&2\\ 2&1&8&1&2 \end{array}\right] \cdot \left[\begin{array} .0& 0& 38& -34& 18\\ 0& 0& 21& 30& -28\\ 0& 0& -15& 8& 20\\ 206& -206& -236& 16& 40\\ 0& 206& -58& -24& -60 \end{array}\right] $
$\left[\begin{array} .a&b&c&d&e\\ f&g&h&i&j\\ k&l&m&n&o\\ p&q&r&s&t\\ u&v&w&x&y \end{array}\right] = \frac{1}{206} \left[\begin{array} .206& 206& -44& -4& -10&\\ 206& 206& -176& -16& -40&\\ 206& 206& -264& -24& -60&\\ 206& 206& -255& -70& -72&\\ 206& 206& -375& -6& 88& \end{array}\right]. $
In a finite number system mod 26 or mod 29, the method is EXACTLY the same just that the computation like matrix/vector multiplications and then the matrix inverse will be different. Side note, in this case the system is breakable in all three cases; the real one (like I showed), mod 26, and mod 29 because the determinant of the plaintext matrix (which we are inverting) is 206 which is nonzero in all three systems.
And no we never assumed you were trying to do something illegal :-) because this method has absolutely zero security in the real world. No one actually uses the Hill cipher for anything useful/valuable. Are you trying to read your sister's diary?