Matrix Representation for logical gates?
Yes, "all of the propositional logical operators can be done with a unique binary 2×2 matrix". 

In fact, every possible map $g: \{0,1\}\times\{0,1\}\to\{0,1\}$ can be done with a unique binary 2x2 matrix. 

Here is a proof.

$$\begin{align}
 &amp;\quad \begin{bmatrix} X, \  \neg X  \end{bmatrix}\begin{bmatrix} A &amp; B\\ C &amp; D\\ \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} \\&amp;= \begin{bmatrix} XA \oplus \neg XC, \ XB \oplus \neg XD \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} \\ 
&amp;= XAY \oplus \neg XCY \oplus XB\neg Y \oplus \neg XD\neg Y\\
\end{align}$$


Suppose the above expression equals $g(X,Y)$. 


Let $X=0, Y=0$, we get $0\oplus0\oplus0\oplus D=D=g(0,0)$.
Let $X=0, Y=1$, we get $0\oplus C\oplus0\oplus 0=C=g(0,1)$.
Let $X=1, Y=0$, we get $0\oplus0\oplus B\oplus 0=B=g(1,0)$.
Let $X=1, Y=1$, we get $A\oplus0\oplus0\oplus 0=A=g(1,1)$.

Conversely, if we have $A= g(1,1)$, $B= g(1,0)$, $C=g(0,1)$ and $D=g(0,0)$, the above expression can be verified to be equal to $g(X,Y)$ for all $X, Y$.


So we have proved that 

$$ \begin{bmatrix} X, \  \neg X  \end{bmatrix}\begin{bmatrix} A &amp; B\\ C &amp; D\\ \end{bmatrix} \begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} = g(X,Y) \\ \Leftrightarrow A= g(1,1),\ B= g(1,0),\ C=g(0,1),\ D=g(0,0)$$

If we use in-fix notation for $g$, we have
$$ \begin{bmatrix} X, \  \neg X  \end{bmatrix}
\begin{bmatrix} 1g1 &amp; 1g0\\ 0g1 &amp; 0g0\\ \end{bmatrix} 
\begin{bmatrix} Y \\ \neg Y \\ \end{bmatrix} = XgY, $$
or
$$ \begin{bmatrix} 1g1 &amp; 1g0\\ 0g1 &amp; 0g0\\ \end{bmatrix} 
 \equiv g. $$
In particular, we can substitute $g$ by anyone of $\land, \lor, \oplus, \odot, \Rightarrow, \Leftarrow, \equiv$.



For more related concepts and computation, please check boolean rings.