1. **State the problem:** We are given matrix $$A = \begin{bmatrix}-6 & -9 & -8 \\ 2 & 9 & 6 \\ 0 & 1 & -1 \end{bmatrix}$$ and asked to find the third row of the adjugate matrix $$\operatorname{adj}(A)$$. 2. **Recall the definition:** The adjugate matrix $$\operatorname{adj}(A)$$ is the transpose of the cofactor matrix of $$A$$. 3. **Step to find the third row of $$\operatorname{adj}(A)$$:** - The third row of $$\operatorname{adj}(A)$$ is the transpose of the third column of the cofactor matrix. - So, find cofactors $$C_{13}, C_{23}, C_{33}$$ (cofactors of elements in the third column of $$A$$). 4. **Calculate cofactors:** - $$C_{13} = (-1)^{1+3} \det \begin{bmatrix} 2 & 9 \\ 0 & 1 \end{bmatrix} = -1 \times (2 \times 1 - 0 \times 9) = -2$$ - $$C_{23} = (-1)^{2+3} \det \begin{bmatrix} -6 & -9 \\ 0 & 1 \end{bmatrix} = 1 \times (-6 \times 1 - 0 \times (-9)) = -6$$ - $$C_{33} = (-1)^{3+3} \det \begin{bmatrix} -6 & -9 \\ 2 & 9 \end{bmatrix} = 1 \times (-6 \times 9 - 2 \times (-9)) = -54 + 18 = -36$$ 5. **Form the third row of $$\operatorname{adj}(A)$$:** $$$$(C_{13}, C_{23}, C_{33}) = (-2, -6, -36)$$ 6. **Check the sign and order:** The third row of $$\operatorname{adj}(A)$$ is the transpose of the third column of cofactors, so it is $$(-2, -6, -36)$$. 7. **Compare with options:** None of the options exactly match $$(-2, -6, -36)$$, but option [2, -6, -36] is close except for the first element's sign. **Final answer:** The third row of $$\operatorname{adj}(A)$$ is $$\boxed{[-2, -6, -36]}$$.