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(D) The characteristic equation is $\lambda^2 - 4\lambda + 3 = 0$, which factors as $(\lambda - 1)(\lambda - 3) = 0$. The eigenvalues are $\lambda_1 =

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$19$

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(D) The characteristic equation is $\lambda^2 - 4\lambda + 3 = 0$, which factors as $(\lambda - 1)(\lambda - 3) = 0$. The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 3$. For any $2 	imes 2$ matrix $A$, the trace is $	ext{tr}(A) = a + d = \lambda_1 + \lambda_2 = 1 + 3 = 4$ and the determinant is $\det(A) = ad - bc = \lambda_1 \lambda_2 = 1 	imes 3 = 3$. We are given $a, d \in \{0, 1, 2, 3, 4\}$ and $a + d = 4$, so $ad - bc = 3 \implies bc = ad - 3$. Case $1$: $(a, d) = (1, 3)$. Then $bc = (1)(3) - 3 = 0$. Possible $(b, c)$ pairs such that $bc = 0$ with $b, c \in \{0, 1, 2, 3, 4\}$: If $b=0$, $c \in \{0, 1, 2, 3, 4\}$ ($5$ values). If $c=0$, $b \in \{0, 1, 2, 3, 4\}$ ($5$ values). Since $(0, 0)$ is counted twice, total pairs = $5 + 5 - 1 = 9$. Case $2$: $(a, d) = (3, 1)$. Then $bc = (3)(1) - 3 = 0$. Similar to Case $1$, total pairs = $9$. Case $3$: $(a, d) = (2, 2)$. Then $bc = (2)(2) - 3 = 1$. Possible $(b, c)$ pairs such that $bc = 1$ with $b, c \in \{0, 1, 2, 3, 4\}$: $(1, 1)$ is the only pair. Total pairs = $1$. Case $4$: $(a, d) = (0, 4)$ or $(4, 0)$. Then $bc = (0)(4) - 3 = -3$. Since $b, c \ge 0$, $bc = -3$ is impossible. Total number of matrices = $9 + 9 + 1 = 19$.

Let $S = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \{0, 1, 2, 3, 4 \} \text{ and } A^2 - 4A + 3I = 0 \right\}$ be a set of $2 \times 2$ matrices. Then the number of matrices in $S$, for which the sum of the diagonal elements is equal to $4$, is:

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