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(D) Let $A = \begin{bmatrix} 2 & 5 & 0 \ 0 & 1 & 1 \ -1 & 0 & 3 \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = 2(1 	imes 3 - 1 	im

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$ \begin{bmatrix} 3 & -15 & 5 \ -1 & 6 & -2 \ 1 & -5 & 2 \end{bmatrix} $

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(D) Let $A = \begin{bmatrix} 2 & 5 & 0 \ 0 & 1 & 1 \ -1 & 0 & 3 \end{bmatrix}$.First,we calculate the determinant $|A|$:$|A| = 2(1 	imes 3 - 1 	imes 0) - 5(0 	imes 3 - 1 	imes (-1)) + 0(0 	imes 0 - 1 	imes (-1))$$|A| = 2(3) - 5(1) + 0 = 6 - 5 = 1$.Since $|A| 
eq 0$,the inverse exists.Next,we find the matrix of cofactors $C_{ij}$:$C_{11} = +(3-0) = 3, C_{12} = -(0+1) = -1, C_{13} = +(0+1) = 1$$C_{21} = -(15-0) = -15, C_{22} = +(6-0) = 6, C_{23} = -(0+5) = -5$$C_{31} = +(5-0) = 5, C_{32} = -(2-0) = -2, C_{33} = +(2-0) = 2$The adjoint matrix $	ext{adj}(A)$ is the transpose of the cofactor matrix:$	ext{adj}(A) = \begin{bmatrix} 3 & -1 & 1 \ -15 & 6 & -5 \ 5 & -2 & 2 \end{bmatrix}^T = \begin{bmatrix} 3 & -15 & 5 \ -1 & 6 & -2 \ 1 & -5 & 2 \end{bmatrix}$.Finally,$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1} \begin{bmatrix} 3 & -15 & 5 \ -1 & 6 & -2 \ 1 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -15 & 5 \ -1 & 6 & -2 \ 1 & -5 & 2 \end{bmatrix}$.Thus,the correct option is $D$.

The inverse of the matrix $ \begin{bmatrix} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{bmatrix} $ is

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