If $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1\end{array}\right], A^{-1}=\frac{1}{2}\left[\begin{array}{ccc}1 & -1 & 1 \\ -8 & 6 & 2 y \\ 5 & -3 & 1\end{array}\right]$, then the point $(x, y)$ lies on the curve represented by the equation.
- A
$y=3 x^2-5 x-1$
- B
$y=\log _{2 / 5}\left(2^x+2^{-x}\right)$
- C
$y=\frac{e^x+1}{e^x-1}$
- D
$3 x^2 y-5 x y+12=0$
✓ Correct answer: B
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