I have searched but an unable to find any examples like what I am faced with.
Plaintext = SOLVED CipherText = GEZXDS 2x2 encryption matrix
$$ \left(\begin{matrix} 11 & 21 \\ 4 & 3 \end{matrix} \right)*\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)= \left(\begin{matrix} 25 & 23 \\ 3 & 18 \end{matrix}\right) $$
with the result: $$11a +21c \equiv 25 \text{(mod 26)}$$ $$4a + 3c \equiv 3 \text{(mod 26)}$$
and
$$11b + 21d \equiv 23 \text{(mod 26)}$$ $$4b + 3d \equiv 18 \text{(mod 26)}$$
I have found a solution. solve for a: $$11a + 21c = 25 (mod 26)$$ $$4a + 3c = 3 (mod 26)$$
-> (multiply second equation by 7) =
$$11a + 21c = 25 (mod 26)$$ $$28a + 21c = 21 (mod 26)$$
-> apply (mod26) where required and Subtract =
$$9a = 4 (mod 26)$$ 9 inverse mod 26 = 3 $$a=3*4 =12$$
Solve for c:
$$11a = 132$$ $$2 + 21c = 25 (mod 26)$$ $$21c = 23$$ 21 inverse = 5 $$c = 5*23 (mod 26)$$ $$c=11$$
and so on.
Thanks for everyone's help
Addressing the question about multiplying by seven.
If you know that the congruence $$ x\equiv y\pmod{26} \tag{1} $$ is true, then you also know that the congruence $$ 7x\equiv 7y\pmod{26}\tag{2} $$ is true. This is because the first congruence simply states that $x-y$ is divisible by $26$. Consequently also $7(x-y)=7x-7y$ is divisible by $26$ which is exactly what the second congruence claims. In other words $(2)$ is a consequence of $(1)$, or $(1)\implies (2)$. The logic will automatically flow in this direction and we could have used any integer in place of $7$ here. The real question is whether we can also go backwards? In other words, does $(2)\implies (1)$ or, can we deduce that $(1)$ is true if and only if $(2)$ is true.
To that end it is necessary that $7$ and $26$ are coprime. Simply stated this is a consequence of the fundamental theorem of arithmetic (=uniqueness of prime factorization of integers): multiplying an integer by seven will not make it divisible by $26$ unless it already is. The same conclusion can be reached by using a modular inverse of $7$. You have hopefully seen how that can be found (when it exists) using the extended Euclidean algorithm, so I simply make the observation that $$ 7\cdot15=105=1+4\cdot26\equiv1\pmod{26}. $$ How does this help? Repeating the above logic we see that because $7\cdot15=105$ the congruence $(2)$ implies the congruence $$ 105x\equiv105y\pmod{26}\tag{3}. $$ But on the left hand side of $(3)$ we have $105x\equiv x$ and on the right hand side we have $105y\equiv y$, because both $104x$ and $104y$ are divisible by $26$ for all integers $x,y$. Thus $(3)\implies(1)$, and we are done.
Addressing the question of finding the key matrix (the question was edited by the OP as it turned out that he had made an unlucky choice of two ciphertex/plaintext pairs)
$$ \left(\begin{matrix} 11 & 21 \\ 4 & 3 \end{matrix} \right)*\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)= \left(\begin{matrix} 25 & 23 \\ 3 & 18 \end{matrix}\right). $$
You can solve this system of congruences any logical way you can/want. I show how to do it using the inverse matrix. The general recipe for inverting a $2\times2$ matrix (hopefully you have seen this in linear algebra or elsewhere) is $$ \left(\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}\right)^{-1} =\frac1{a_{11}a_{22}-a_{12}a_{21}} \left(\begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{matrix}\right).\tag{4} $$ Everything here makes sense in modulo $26$ arithmetic except possibly division by the determinant. As above, we can divide by an integer modulo $26$ by multiplying with the modular inverse provided that the inverse exists, i.e. provided that the number we divide by is coprime to $26$.
The determinant of your matrix $$ A=\left(\begin{matrix} 11 & 21 \\ 4 & 3 \end{matrix} \right) $$ is $\det A=33-84=-51\equiv-51+2\cdot26=1\pmod{26},$ which is, of course, equal to its own inverse. Therefore $$ A^{-1}=\left(\begin{matrix} 11 & 21 \\ 4 & 3 \end{matrix} \right)^{-1}=\frac11\left(\begin{matrix} 3 & -21 \\ -4 & 11 \end{matrix} \right)\equiv \left(\begin{matrix} 3 & 5 \\ 22 & 11 \end{matrix} \right)\pmod{26}, $$ as $-21\equiv5\pmod{26}$ et cetera. We can then solve the matrix equation $(4)$ simply by multiplying it by $A^{-1}$ from the left hand side. The result is $$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)= \left(\begin{matrix} 3 & 5 \\ 22 & 11 \end{matrix} \right) \left(\begin{matrix} 25 & 23 \\ 3 & 18 \end{matrix}\right)= \left(\begin{matrix} 90 & 159 \\ 583 & 704 \end{matrix}\right) \equiv \left(\begin{matrix} 12 & 3 \\ 11 & 2 \end{matrix}\right)\pmod{26}. $$
Leaving it to you to check all the calculations and the verification that this key matrix is compatible with the third plaintext/ciphertext pair that you have.