1. **State the problem:** We need to find matrix $A$ such that $$\left(3A + \begin{bmatrix}-1 & -2 \\ -5 & 4 \\ 2 & 4\end{bmatrix}\right)^T = \begin{bmatrix}-4 & 2 & 5 \\ -3 & 3 & -1\end{bmatrix}.$$ 2. **Understand the dimensions:** Matrix inside parentheses is $3 \times 2$, so $3A$ must also be $3 \times 2$, meaning $A$ is $3 \times 2$. 3. **Transpose operation:** Transpose of a $3 \times 2$ matrix is $2 \times 3$, matching the right side. 4. **Rewrite the equation:** $$\left(3A + M\right)^T = N,$$ where $$M = \begin{bmatrix}-1 & -2 \\ -5 & 4 \\ 2 & 4\end{bmatrix}, \quad N = \begin{bmatrix}-4 & 2 & 5 \\ -3 & 3 & -1\end{bmatrix}.$$ 5. **Transpose both sides:** Taking transpose again, $$3A + M = N^T.$$ 6. **Calculate $N^T$:** $$N^T = \begin{bmatrix}-4 & -3 \\ 2 & 3 \\ 5 & -1\end{bmatrix}.$$ 7. **Solve for $A$:** $$3A = N^T - M = \begin{bmatrix}-4 & -3 \\ 2 & 3 \\ 5 & -1\end{bmatrix} - \begin{bmatrix}-1 & -2 \\ -5 & 4 \\ 2 & 4\end{bmatrix} = \begin{bmatrix}-4 - (-1) & -3 - (-2) \\ 2 - (-5) & 3 - 4 \\ 5 - 2 & -1 - 4\end{bmatrix} = \begin{bmatrix}-3 & -1 \\ 7 & -1 \\ 3 & -5\end{bmatrix}.$$ 8. **Divide both sides by 3:** $$A = \frac{1}{3} \begin{bmatrix}-3 & -1 \\ 7 & -1 \\ 3 & -5\end{bmatrix} = \begin{bmatrix}\cancel{\frac{-3}{3}} & \frac{-1}{3} \\ \frac{7}{3} & \cancel{\frac{-1}{3}} \\ \cancel{\frac{3}{3}} & \frac{-5}{3}\end{bmatrix} = \begin{bmatrix}-1 & -\frac{1}{3} \\ \frac{7}{3} & -\frac{1}{3} \\ 1 & -\frac{5}{3}\end{bmatrix}.$$ **Final answer:** $$A = \begin{bmatrix}-1 & -\frac{1}{3} \\ \frac{7}{3} & -\frac{1}{3} \\ 1 & -\frac{5}{3}\end{bmatrix}.$$