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

8

The correct answer is Option (4) → 8

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Given:

$A = \begin{bmatrix} 0 & 1 & 3 \\ 1 & 2 & x \\ 2 & 3 & 1 \end{bmatrix}, \;\; A^{-1} = \begin{bmatrix} \frac{1}{2} & -4 & \frac{5}{2} \\[6pt] -\frac{1}{2} & 3 & -\frac{3}{2} \\[6pt] \frac{1}{2} & 3 & \frac{1}{2} \end{bmatrix}$

Condition: $A \cdot A^{-1} = I$

Second row of $A$ with third column of $A^{-1}$:

$[1 \;\; 2 \;\; x] \cdot \begin{bmatrix} \frac{5}{2} \\ -\frac{3}{2} \\ \frac{1}{2} \end{bmatrix}$

$= \; 1 \cdot \frac{5}{2} + 2 \cdot \left(-\frac{3}{2}\right) + x \cdot \frac{1}{2}$

$= \frac{5}{2} - 3 + \frac{x}{2}$

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

In identity matrix, this element $= 0$.

So, $\frac{x}{2} - \frac{1}{2} = 0 \;\;\Rightarrow\;\; x = 1$

Therefore, $8x = 8 \times 1 = 8$

Final Answer: 8