Let A be a 3 × 3 matrix such that A^T [1; 0; 1] = [5; 2; 2], A^T [0; 0; 1] = [3; 1; 1], A [1; 0; 1] = [3; 4; 4] and A [0; 0; 1] = [1; 3; 1]. If det (…

Let $\displaystyle A$ be a $\displaystyle 3 \times 3$ matrix such that $$\mathrm{A}^{\mathrm{T}}\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 5 \\ 2 \\ 2 \end{array}\right], \mathrm{A}^{\mathrm{T}}\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 1 \\ 1 \end{array}\right], \mathrm{A}\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \\ 4 \end{array}\right] \text { and } \mathrm{A}\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 1 \\ 3 \\ 1 \end{array}\right] . $$ If $\displaystyle \operatorname{det}(\mathrm{A})=1$, then $\displaystyle \operatorname{det}\left(\operatorname{adj}\left(\mathrm{A}^2+\mathrm{A}\right)\right)$ is equal to: