If for any $2 \times 2$ square matrix $A$, $A(\operatorname{adj} A)=\left[\begin{array}{ll}8 & 0 \\ 0 & 8\end{array}\right]$, then the value of $\operatorname{det}(A) .$

For any square matrix $A$ of order $n$, the fundamental property involving the adjoint matrix is $A(\operatorname{adj} A) = (\operatorname{det} A) I_n$, where $I_n$ is the identity matrix of order $n$.

Given that $A$ is a $2 \times 2$ matrix, $n = 2$. Therefore, $I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

The given equation is $A(\operatorname{adj} A) = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix}$.

Substituting the property into the given equation, we have $(\operatorname{det} A) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix}$.

Comparing the matrices, we get $(\operatorname{det} A) = 8$.

Answer: 8