1. **State the problem:** We need to find the determinant of the matrix $$A = \begin{bmatrix} 0 & 1 & 5 \\ 3 & -6 & 9 \\ 2 & 6 & 1 \end{bmatrix}$$ using row reduction. 2. **Recall the determinant properties with row operations:** - Swapping two rows multiplies the determinant by $$-1$$. - Multiplying a row by a scalar $$k$$ multiplies the determinant by $$k$$. - Adding a multiple of one row to another does not change the determinant. 3. **Apply row operations to get an upper triangular matrix:** Start with $$A$$: $$\begin{bmatrix} 0 & 1 & 5 \\ 3 & -6 & 9 \\ 2 & 6 & 1 \end{bmatrix}$$ Since the first element in row 1 is zero, swap row 1 and row 2 to get a nonzero pivot: $$R_1 \leftrightarrow R_2$$ $$\Rightarrow \begin{bmatrix} 3 & -6 & 9 \\ 0 & 1 & 5 \\ 2 & 6 & 1 \end{bmatrix}$$ Determinant changes sign: $$\det = -\det(A)$$ 4. **Eliminate below the pivot in column 1:** Use $$R_3 \to R_3 - \frac{2}{3} R_1$$: $$R_3 = \begin{bmatrix} 2 & 6 & 1 \end{bmatrix} - \frac{2}{3} \times \begin{bmatrix} 3 & -6 & 9 \end{bmatrix} = \begin{bmatrix} 2 - 2, 6 - (-4), 1 - 6 \end{bmatrix} = \begin{bmatrix} 0 & 10 & -5 \end{bmatrix}$$ Matrix now: $$\begin{bmatrix} 3 & -6 & 9 \\ 0 & 1 & 5 \\ 0 & 10 & -5 \end{bmatrix}$$ 5. **Eliminate below the pivot in column 2:** Use $$R_3 \to R_3 - 10 R_2$$: $$R_3 = \begin{bmatrix} 0 & 10 & -5 \end{bmatrix} - 10 \times \begin{bmatrix} 0 & 1 & 5 \end{bmatrix} = \begin{bmatrix} 0, 10 - 10, -5 - 50 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -55 \end{bmatrix}$$ Matrix now: $$\begin{bmatrix} 3 & -6 & 9 \\ 0 & 1 & 5 \\ 0 & 0 & -55 \end{bmatrix}$$ 6. **Calculate determinant:** The matrix is now upper triangular, so the determinant is the product of the diagonal entries times the sign change from the row swap: $$\det(A) = - (3 \times 1 \times -55) = - ( -165 ) = 165$$ **Final answer:** $$\boxed{165}$$