Let $A=\left[a_{i j} ight]$ be a matrix of order $3 imes 3$, 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 :

Question

Let $A=\left[a_{i j}ight]$ be a matrix of order $3 	imes 3$, 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 :

Aicharya · Accepted Answer

$\begin{aligned} & A=\left[\begin{array}{lll}(\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6\end{array}\right] \\ & A=\left[\begin{array}{ccc}2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8\end{array}\right] \\ & A^2=2^2\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]\end{aligned}$
$=4\left[\begin{array}{ccc}- & - & - \\ - & - & - \\ (2+4+8) & (2 \sqrt{2}+4 \sqrt{2}+8 \sqrt{2}) & (4+8+16)\end{array}\right]$

Sum of elements of $3^{\text {nd }}$ row $=4(14+14 \sqrt{2}+28)$
$\begin{aligned}
& =4(42+14 \sqrt{2}) \\ & =168+56 \sqrt{2} \\ & \alpha+\beta \sqrt{2} \\ \therefore & \alpha+\beta=168+56=224
\end{aligned}$

Let $A=\left[a_{i j}\right]$ be a matrix of order $3 \times 3$, with $a_{i j}=(\sqrt{2})^{i+j}$. If the sum…

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