1. **Problem:** Given matrices $$A=\begin{bmatrix}-2 & 1 & \frac{3}{2} \\ 0 & 3 & 2\end{bmatrix},\quad B=\begin{bmatrix}1 & 2 & 3 \\ 3 & 2 & 1\end{bmatrix}$$ Find the matrix $$2A + 3B$$. 2. **Formula:** To add or multiply matrices by scalars, multiply each element by the scalar and then add corresponding elements: $$2A = 2 \times A, \quad 3B = 3 \times B$$ $$2A + 3B = \begin{bmatrix}2a_{11} & 2a_{12} & 2a_{13} \\ 2a_{21} & 2a_{22} & 2a_{23}\end{bmatrix} + \begin{bmatrix}3b_{11} & 3b_{12} & 3b_{13} \\ 3b_{21} & 3b_{22} & 3b_{23}\end{bmatrix}$$ 3. **Calculate:** $$2A = \begin{bmatrix}2 \times (-2) & 2 \times 1 & 2 \times \frac{3}{2} \\ 2 \times 0 & 2 \times 3 & 2 \times 2\end{bmatrix} = \begin{bmatrix}-4 & 2 & 3 \\ 0 & 6 & 4\end{bmatrix}$$ $$3B = \begin{bmatrix}3 \times 1 & 3 \times 2 & 3 \times 3 \\ 3 \times 3 & 3 \times 2 & 3 \times 1\end{bmatrix} = \begin{bmatrix}3 & 6 & 9 \\ 9 & 6 & 3\end{bmatrix}$$ 4. **Add matrices:** $$2A + 3B = \begin{bmatrix}-4+3 & 2+6 & 3+9 \\ 0+9 & 6+6 & 4+3\end{bmatrix} = \begin{bmatrix}-1 & 8 & 12 \\ 9 & 12 & 7\end{bmatrix}$$ 5. **Compare with options:** The closest option is $$D = \begin{bmatrix}-1 & 8 & 15 \\ 9 & 12 & 7\end{bmatrix}$$ Note the difference in the element at position (1,3) which is 12 in our calculation but 15 in option D. Since none exactly match, the closest is option D. **Final answer:** Option D --- **Summary:** The value of $$2A + 3B$$ is approximately $$\begin{bmatrix}-1 & 8 & 12 \\ 9 & 12 & 7\end{bmatrix}$$, matching option D most closely.