If $A = \begin{bmatrix} 0 & 1 & 3 \\ 1 & 2 & x \\ 2 & 3 & 1 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \frac{1}{2} & -4 & \frac{5}{2} \\ -\frac{1}{2} & 3 & -\frac{3}{2} \\ \frac{1}{2} & y & \frac{1}{2} \end{bmatrix}$, then the value of $8x + 5y$ is:

1. Solve for $x$

We know that the product of the second row of $A$ and the first column of $A^{-1}$ must equal the element at $(2,1)$ in the identity matrix, which is 0.

$[1 \quad 2 \quad x] \begin{bmatrix} 1/2 \\ -1/2 \\ 1/2 \end{bmatrix} = 0$

$(1 \cdot \frac{1}{2}) + (2 \cdot -\frac{1}{2}) + (x \cdot \frac{1}{2}) = 0$

$\frac{1}{2} - 1 + \frac{x}{2} = 0$

$-\frac{1}{2} + \frac{x}{2} = 0$

$\frac{x}{2} = \frac{1}{2} \implies \mathbf{x = 1}$

2. Solve for $y$

Now, we look at the product of the first row of $A$ and the second column of $A^{-1}$, which must equal the element at $(1,2)$ in the identity matrix, which is also 0.

$[0 \quad 1 \quad 3] \begin{bmatrix} -4 \\ 3 \\ y \end{bmatrix} = 0$

$(0 \cdot -4) + (1 \cdot 3) + (3 \cdot y) = 0$

$0 + 3 + 3y = 0$

$3y = -3 \implies \mathbf{y = -1}$

3. Calculate the Final Value

Now we substitute the values of $x$ and $y$ into the expression $8x + 5y$:

$8(1) + 5(-1)$

$8 - 5 = \mathbf{3}$

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Final Answer:

The value of $8x + 5y$ is 3. The correct option is 4.