Determinant Adjoint 643519

1. **State the problem:** We need to find the determinant of the adjoint of the matrix $$A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 3 & 2 \\ 0 & 4 & 5 \end{pmatrix}$$. 2. **Recall the formula:** The adjoint of a matrix $A$, denoted $\operatorname{adj}(A)$, satisfies $$\det(\operatorname{adj}(A)) = (\det(A))^{n-1}$$ where $n$ is the size of the square matrix. Here, $n=3$. 3. **Calculate $\det(A)$:** $$\det(A) = 2 \times \begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix} - 1 \times \begin{vmatrix} 1 & 2 \\ 0 & 5 \end{vmatrix} + 0 \times \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix} = 3 \times 5 - 2 \times 4 = 15 - 8 = 7$$ $$\begin{vmatrix} 1 & 2 \\ 0 & 5 \end{vmatrix} = 1 \times 5 - 2 \times 0 = 5$$ So, $$\det(A) = 2 \times 7 - 1 \times 5 + 0 = 14 - 5 = 9$$ 4. **Apply the formula for the determinant of the adjoint:** Since $n=3$, $$\det(\operatorname{adj}(A)) = (\det(A))^{3-1} = (9)^2 = 81$$ 5. **Final answer:** $$\boxed{81}$$