1. **State the problem:** We are given matrix $$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$ with $$\det(A) = 4$$. We need to find the determinant of matrix $$B = \begin{bmatrix} a + 5g & b + 5h & c + 5i \\ g & h & i \\ 6d & 6e & 6f \end{bmatrix}$$. 2. **Recall determinant properties:** - Adding a multiple of one row to another does not change the determinant. - Multiplying a row by a scalar multiplies the determinant by that scalar. - The determinant of a matrix with rows $$R_1, R_2, R_3$$ is denoted $$\det(R_1, R_2, R_3)$$. 3. **Analyze matrix B:** - The first row of $$B$$ is $$R_1^B = R_1^A + 5 R_3^A$$ (since $$a+5g = a + 5g$$, etc.). - The second row of $$B$$ is $$R_2^B = R_3^A$$. - The third row of $$B$$ is $$R_3^B = 6 R_2^A$$. 4. **Use determinant linearity:** Since $$R_1^B = R_1^A + 5 R_3^A$$, by linearity, $$\det(B) = \det(R_1^A + 5 R_3^A, R_3^A, 6 R_2^A) = \det(R_1^A, R_3^A, 6 R_2^A) + 5 \det(R_3^A, R_3^A, 6 R_2^A)$$. 5. **Simplify terms:** - $$\det(R_3^A, R_3^A, 6 R_2^A) = 0$$ because two rows are identical. - So, $$\det(B) = \det(R_1^A, R_3^A, 6 R_2^A)$$. 6. **Rearrange rows to match $$A$$:** We want to express $$\det(R_1^A, R_3^A, 6 R_2^A)$$ in terms of $$\det(A) = \det(R_1^A, R_2^A, R_3^A)$$. Swapping rows changes the sign: $$\det(R_1^A, R_3^A, 6 R_2^A) = - \det(R_1^A, 6 R_2^A, R_3^A)$$ (swap rows 2 and 3). 7. **Factor out scalar from row 2:** $$\det(R_1^A, 6 R_2^A, R_3^A) = 6 \det(R_1^A, R_2^A, R_3^A) = 6 \times 4 = 24$$. 8. **Apply sign from row swap:** $$\det(B) = -24$$. **Final answer:** $$\boxed{-24}$$.