Let $\mathrm{A}=\left[a_{i j}\right]$ be a matrix of order $3 \times 3$, with $a_{i j}=(\sqrt{2})^{i+j}$. If the sum of all the elements in the third row of $A^2$ is $\alpha+\beta \sqrt{2}, \alpha, \beta \in \mathbf{Z}$, then $\alpha+\beta$ is equal to :

<p>$$\begin{aligned} & A=\left[\begin{array}{lll} (\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6 \end{array}\right] \\ & A=\left[\begin{array}{ccc} 2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8 \end{array}\right] \\ & A^2=2^2\left[\begin{array}{ccc} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{array}\right]\left[\begin{array}{ccc} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{array}\right] \end{aligned}$$</p> <p>$$\begin{aligned} &=4\left[\begin{array}{ccc} - & - & - \\ - & - & - \\ (2+4+8) & (2 \sqrt{2}+4 \sqrt{2}+8 \sqrt{2}) & (4+8+16) \end{array}\right]\\ &\text { Sum of elements of } 3^{\text {rd }} \text { row }=4(14+14 \sqrt{2}+28)\\ &\begin{aligned} & =4(42+14 \sqrt{2}) \\ & =168+56 \sqrt{2} \\ & \alpha+\beta \sqrt{2} \\ \therefore \quad & \alpha+\beta=168+56=224 \end{aligned} \end{aligned}$$</p>