Number of Matrices Satisfying Matrix Equation and Trace Condition

To find the number of matrices $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ in $S$ satisfying the given conditions, we process the matrix equation and trace condition step-by-step.

Step 1: Simplify the Matrix Equation

We are given that $a, b, c, d \in \{0, 1, 2, 3, 4\}$, the sum of the diagonal elements (trace) is: $\text{tr}(A) = a + d = 4 \implies d = 4 - a$

Now, compute $A^2$: $A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & b(a+d) \\ c(a+d) & d^2 + bc \end{bmatrix}$

Since $a + d = 4$, this simplifies to: $A^2 = \begin{bmatrix} a^2 + bc & 4b \\ 4c & d^2 + bc \end{bmatrix}$

Substitute $A^2$, $A$, and $I$ into the matrix equation $A^2 - 4A + 3I = 0$: $\begin{bmatrix} a^2 + bc & 4b \\ 4c & d^2 + bc \end{bmatrix} - 4\begin{bmatrix} a & b \\ c & d \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Equating the corresponding entries:

1. Off-diagonal elements: $4b - 4b = 0$ and $4c - 4c = 0$ (holds for all $b, c$).
2. Top-left diagonal element: $a^2 + bc - 4a + 3 = 0 \implies bc = 4a - a^2 - 3 = a(4-a) - 3 = ad - 3$
3. Bottom-right diagonal element: $d^2 + bc - 4d + 3 = 0 \implies bc = 4d - d^2 - 3 = d(4-d) - 3 = ad - 3$

Hence, the given conditions reduce to: $ad - bc = 3 \quad \text{and} \quad a + d = 4$

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Step 2: Cases for $(a, d)$

Since $a, d \in \{0, 1, 2, 3, 4\}$ and $a + d = 4$, we analyze all possible pairs $(a, d)$:

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

2. Case 2: $a = 4, d = 0$ $bc = ad - 3 = (4)(0) - 3 = -3$ Similarly, there are no solutions.

3. Case 3: $a = 1, d = 3$ $bc = ad - 3 = (1)(3) - 3 = 0$ We need $b \cdot c = 0$ where $b, c \in \{0, 1, 2, 3, 4\}$:

   • If $b = 0$, then $c \in \{0, 1, 2, 3, 4\}$ (5 possibilities).
   • If $b \in \{1, 2, 3, 4\}$, then $c = 0$ (4 possibilities).

   Total pairs for $(b, c) = 5 + 4 = 9$.

4. Case 4: $a = 3, d = 1$ $bc = ad - 3 = (3)(1) - 3 = 0$ By symmetry with Case 3, there are 9 possibilities for $(b, c)$.

5. Case 5: $a = 2, d = 2$ $bc = ad - 3 = (2)(2) - 3 = 1$ Since $b, c \in \{0, 1, 2, 3, 4\}$, $bc = 1$ gives only 1 possibility: $b = 1, c = 1$.

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Step 3: Total Count

Summing up the valid matrices from all cases: $\text{Total number of matrices} = 0 + 0 + 9 + 9 + 1 = 19$

Thus, the correct option is D.