If $A = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{pmatrix}$, $A^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -1 & 1 \\ -8 & 6 & 2y \\ 5 & -3 & 1 \end{pmatrix}$ then the point $(x,y)$ lies on the curve
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Whenever a matrix inverse is given, use $AA^{-1} = I$ to form equations. Multiply the matrices and compare each element with the identity matrix to solve for unknowns.