The problem asks for the value of $k$ given that the matrix $A = \begin{pmatrix} k & 1 \\ 3 & 2k \end{pmatrix}$ has a determinant of 15. Let's break it down:

1. The determinant of matrix $A$ is calculated as: $\text{det}(A) = (k \times 2k) - (1 \times 3) = 2k^2 - 3$ We're told that the determinant is 15, so: $2k^2 - 3 = 15$
2. Solve for $k$: $2k^2 = 15 + 3 = 18$ $k^2 = \frac{18}{2} = 9$ $k = \sqrt{9} = 3 \quad \text{or} \quad k = -3$

Since the problem specifies that $k > 0$, the solution is $k = 3$.

Thus, the correct answer is A. 3.

Do you need further explanation on any part of this solution?

Here are some related questions to explore further:

1. How do you calculate the determinant of a 2x2 matrix?
2. What would be the result if $k$ were negative in this problem?
3. How does the determinant help in understanding matrix properties?
4. What is the role of the determinant in determining matrix invertibility?
5. What other mathematical properties are influenced by the determinant of a matrix?

Tip: Always double-check the sign and constraints (such as $k > 0$) when solving quadratic equations.