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(B) We know that $A \cdot A^{-1} = I$, where $I$ is the identity matrix of order $3 	imes 3$. Given $A = \begin{bmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3

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$y=\log_{2/5}(2^x+2^{-x})$

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(B) We know that $A \cdot A^{-1} = I$, where $I$ is the identity matrix of order $3 	imes 3$. Given $A = \begin{bmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & x & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \ -8 & 6 & 2y \ 5 & -3 & 1 \end{bmatrix}$. Multiplying $A$ and $A^{-1}$: $A \cdot A^{-1} = \frac{1}{2} \begin{bmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \ -8 & 6 & 2y \ 5 & -3 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$. Considering the element at row $2$, column $3$ of the product matrix: $\frac{1}{2} [ (1)(1) + (2)(2y) + (3)(1) ] = 0$. $1 + 4y + 3 = 0 \implies 4y + 4 = 0 \implies y = -1$. Considering the element at row $3$, column $2$ of the product matrix: $\frac{1}{2} [ (3)(-1) + (x)(6) + (1)(-3) ] = 0$. $-3 + 6x - 3 = 0 \implies 6x - 6 = 0 \implies x = 1$. Thus, the point is $(x, y) = (1, -1)$. Checking the options: For option $B$: $y = \log_{2/5}(2^1 + 2^{-1}) = \log_{2/5}(2 + 0.5) = \log_{2/5}(2.5) = \log_{2/5}(5/2) = -1$. Since $y = -1$ satisfies the equation in option $B$, the point $(1, -1)$ lies on the curve $y = \log_{2/5}(2^x + 2^{-x})$.

If $A=\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{bmatrix}$ and $A^{-1}=\frac{1}{2}\begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & 2y \\ 5 & -3 & 1 \end{bmatrix}$, then the point $(x, y)$ lies on the curve represented by the equation:

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