Let A= [a_i j ] be a matrix of order 3 × 3, with a_i j=(√2)^i+j. If the sum of all the elements in the third row of A^2 is α+β √2, α, β ∈ Z, then α+β…

$\begin{aligned} & A=\left[\begin{array}{lll}(\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6\end{array}\right] \\ & A=\left[\begin{array}{ccc}2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8\end{array}\right] \\ & A^2=2^2\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]\end{aligned}$​A=​(2​)2(2​)3(2​)4​(2​)3(2​)4(2​)5​(2​)4(2​)5(2​)6​​A=​222​4​22​442​​442​8​​A2=22​12​2​2​222​​222​4​​​12​2​2​222​​222​4​​​ $=4\left[\begin{array}{ccc}- & - & - \\ - & - & - \\ (2+4+8) & (2 \sqrt{2}+4 \sqrt{2}+8 \sqrt{2}) & (4+8+16)\end{array}\right]$=4​−−(2+4+8)​−−(22​+42​+82​)​−−(4+8+16)​​ Sum of elements of $3^{\text {nd }}$3nd  row $=4(14+14 \sqrt{2}+28)$=4(14+142​+28) $\begin{aligned}
& =4(42+14 \sqrt{2}) \
& =168+56 \sqrt{2} \
& \alpha+\beta \sqrt{2} \
\therefore & \alpha+\beta=168+56=224
\end{aligned}$