Let A be a 3 × 3 matrix such that A^T bmatrix 1 0 1 bmatrix =...

Let $C_1, C_2, C_3$ be the columns of $A$ and $R_1, R_2, R_3$ be the rows of $A$.

From the given equations, we have:
$A \begin{bmatrix}0\\0\\1\end{bmatrix} = C_3 = \begin{bmatrix}1\\3\\1\end{bmatrix}$

$A \begin{bmatrix}1\\0\\1\end{bmatrix} = C_1 + C_3 = \begin{bmatrix}3\\4\\4\end{bmatrix} \Rightarrow C_1 = \begin{bmatrix}2\\1\\3\end{bmatrix}$

Similarly, for the transpose $A^T$, the columns correspond to the rows of $A$:
$A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = R_3^T = \begin{bmatrix}3\\1\\1\end{bmatrix} \Rightarrow R_3 = \begin{bmatrix}3 & 1 & 1\end{bmatrix}$

$A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = R_1^T + R_3^T = \begin{bmatrix}5\\2\\2\end{bmatrix} \Rightarrow R_1^T = \begin{bmatrix}2\\1\\1\end{bmatrix} \Rightarrow R_1 = \begin{bmatrix}2 & 1 & 1\end{bmatrix}$

Using the elements of $C_1, C_3, R_1,$ and $R_3$, we can construct the matrix $A$ with an unknown central element $b$:
$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & b & 3 \\ 3 & 1 & 1 \end{bmatrix}$

We are given that $\det(A) = 1$. Expanding along the first row:
$\det(A) = 2(b - 3) - 1(1 - 9) + 1(1 - 3b) = 1$

$2b - 6 + 8 + 1 - 3b = 1$

$3 - b = 1 \Rightarrow b = 2$

Thus, the matrix $A$ is:
$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$

We need to find $\det(\operatorname{adj}(A^2 + A))$. Using the properties of determinants and adjoints:
$\det(\operatorname{adj}(A^2 + A)) = (\det(A^2 + A))^2 = (\det(A) \cdot \det(A + I))^2$

First, find $A + I$:
$A + I = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & 3 \\ 3 & 1 & 2 \end{bmatrix}$

Now, calculate $\det(A + I)$:
$\det(A + I) = 3(6 - 3) - 1(2 - 9) + 1(1 - 9)$

$\det(A + I) = 3(3) - 1(-7) + 1(-8) = 9 + 7 - 8 = 8$

Since $\det(A) = 1$, we have:
$\det(A^2 + A) = 1 \times 8 = 8$

Finally, for a $3 \times 3$ matrix:
$\det(\operatorname{adj}(A^2 + A)) = 8^{3-1} = 8^2 = 64$

Answer: $64$