Let S = \ A = bmatrix a & b c & d bmatrix : a, b, c, d ∈ \ 0, 1, 2,...

The characteristic equation of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by:

$A^2 - \text{Tr}(A)A + \det(A)I = 0$

We are given that the sum of the diagonal elements is $4$, so $\text{Tr}(A) = a + d = 4$. Substituting this into the characteristic equation, we get:

$A^2 - 4A + (ad - bc)I = 0$

We are also given that the matrix satisfies:

$A^2 - 4A + 3I = 0$

Comparing the two equations, we must have:

$ad - bc = 3$

We need to find the number of quadruples $(a, b, c, d)$ with elements from the set $\{0, 1, 2, 3, 4\}$ such that $a + d = 4$ and $ad - bc = 3$. Let us analyze the possible values for $a$ and $d$:

Case 1: $a = 0, d = 4$
$ad - bc = 3 \implies 0 - bc = 3 \implies bc = -3$
Since $b, c \in \{0, 1, 2, 3, 4\}$, $bc \ge 0$. Thus, there are no solutions.

Case 2: $a = 1, d = 3$
$ad - bc = 3 \implies 3 - bc = 3 \implies bc = 0$
For $bc = 0$, either $b = 0$ or $c = 0$.
If $b = 0$, $c$ can take any of the $5$ values from the set.
If $c = 0$, $b$ can take any of the $5$ values from the set.
The pair $(0, 0)$ is counted twice, so the number of pairs $(b, c)$ is $5 + 5 - 1 = 9$.

Case 3: $a = 2, d = 2$
$ad - bc = 3 \implies 4 - bc = 3 \implies bc = 1$
The only possible pair from the given set is $b = 1, c = 1$. This gives $1$ solution.

Case 4: $a = 3, d = 1$
$ad - bc = 3 \implies 3 - bc = 3 \implies bc = 0$
Similar to Case 2, there are $9$ possible pairs for $(b, c)$.

Case 5: $a = 4, d = 0$
$ad - bc = 3 \implies 0 - bc = 3 \implies bc = -3$
Again, there are no solutions.

Adding the valid matrices from all cases, the total number of matrices is $9 + 1 + 9 = 19$.

Answer: $19$