For any $n \times n$ matrix, the product of its eigenvalues is equal to the determinant of the matrix:
\[ \lambda_1 \lambda_2 \dots \lambda_n = \text{det}(A) \]
For a $2 \times 2$ matrix, if one eigenvalue is negative and the other is positive (which is usually the case when the trace is positive), the product of the eigenvalues must be negative. Therefore, $\text{det}(A) < 0$.
Given the matrix $A = \begin{bmatrix} 1 & a \\ 8 & 3 \end{bmatrix}$ where $a > 0$.
The trace of the matrix is $\text{Tr}(A) = 1 + 3 = 4$. Since the sum of eigenvalues is positive ($ \lambda_1 + \lambda_2 = 4 $), for at least one eigenvalue to be negative, the product of eigenvalues must be negative:
\[ \lambda_1 \lambda_2 < 0 \implies \text{det}(A) < 0 \]
Calculate the determinant:
\[ \text{det}(A) = (1 \times 3) - (8 \times a) \]
\[ 3 - 8a < 0 \]
\[ 3 < 8a \implies a > \frac{3}{8} \]
$a$ must be greater than $\frac{3}{8}$ (Option A).
