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If $\begin{bmatrix} x & y^3 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 8 \\ 2 & 0 \end{bmatrix}$, then $\begin{bmatrix} x&y\\ 2 & 0 \end{bmatrix}^{-1}$ is equal to :


( A ) $\begin{bmatrix} 0 & -2\\ -2 & 1 \end{bmatrix}$
( B ) $\begin{bmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{4} \end{bmatrix}$
( C ) $\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
( D ) $\begin{bmatrix} 0 & -8\\ -2 & 1 \end{bmatrix}$

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