Let A = [a_ij] be a matrix of order 3 × 3, with ${\mathrm{a}}_{\mathrm{ij}}=(\sqrt{2}{)}^{\mathrm{i}+\mathrm{j}}$. If the sum of all the elements in the
third row of A² is $\mathrm{\alpha }+\mathrm{\beta }\sqrt{2},\mathrm{\alpha },\mathrm{\beta }\in \mathbf{Z}, \mathrm{then} \mathrm{\alpha }+\mathrm{\beta }$ is equal to

$$\begin{gathered} \begin{array}{l}\mathrm{A}=\left[\begin{array}{lll}(\sqrt{2}{)}^2& (\sqrt{2}{)}^3& (\sqrt{2}{)}^4\\ (\sqrt{2}{)}^3& (\sqrt{2}{)}^4& (\sqrt{2}{)}^5\\ (\sqrt{2}{)}^4& (\sqrt{2}{)}^5& (\sqrt{2}{)}^6\end{array}\right]\\ \mathrm{A}=\left[\begin{array}{ccc}2& 2\sqrt{2}& 4\\ 2\sqrt{2}& 4& 4\sqrt{2}\\ 4& 4\sqrt{2}& 8\end{array}\right]\\ {\mathrm{A}}^2=2^2\left[\begin{array}{ccc}1& \sqrt{2}& 2\\ \sqrt{2}& 2& 2\sqrt{2}\\ 2& 2\sqrt{2}& 4\end{array}\right]\left[\begin{array}{ccc}1& \sqrt{2}& 2\\ \sqrt{2}& 2& 2\sqrt{2}\\ 2& 2\sqrt{2}& 4\end{array}\right]\end{array} \\ =4\left[\begin{array}{ccc}-& -& -\\ -& -& -\\ (2+4+8)& (2\sqrt{2}+4\sqrt{2}+8\sqrt{2})& (4+8+16)\end{array}\right] \end{gathered}$$
sum of elements of 3^rd row = $4(14+14\sqrt{2}+28)$
$\begin{array}{l}=4(42+14\sqrt{2})\\ =168+56\sqrt{2}\\ \mathrm{\alpha }+\mathrm{\beta }\sqrt{2}\\ \therefore \mathrm{\alpha }+\mathrm{\beta }=168+56=224\end{array}$