Let $A=\left[{a}_{ij}\right]$ 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 $\alpha+\beta \sqrt{2}, \alpha, \beta \in Z$, then $\alpha+\beta$ is equal to : [JEE Main 2025, 29 Jan (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: b) 224. $A=\left[\begin{matrix}{(\sqrt{2})}^{2} & {(\sqrt{2})}^{3} & {(\sqrt{2})}^{4} \ {(\sqrt{2})}^{3} & {(\sqrt{2})}^{4} & {(\sqrt{2})}^{5} \ {(\sqrt{2})}^{4} & {(\sqrt{2})}^{5} & {(\sqrt{2})}^{6}\end{matrix}\right]$ $A=\left[\begin{matrix}2 & 2\sqrt{2} & 4 \ 2\sqrt{2} & 4 & 4\sqrt{2} \ 4 & 4\sqrt{2} & 8\end{matrix}\right]$ ${A}^{2}={2}^{2}\left[\begin{matrix}1 & \sqrt{2} & 2 \ \sqrt{2} & 2 & 2\sqrt{2} \ 2 & 2\sqrt{2} & 4\end{matrix}\right]\left[\begin{matrix}1 & \sqrt{2} & 2 \ \sqrt{2} & 2 & 2\sqrt{2} \ 2 & 2\sqrt{2} & 4\end{matrix}\right]$ $=4\left[\begin{matrix}− & − & − \ − & − & − \ (14) & (14\sqrt{2}) & (28)\end{matrix}\right]$ Sum of the all elements of third row $=4(14+14\sqrt{2}+28)\ =4(42+14\sqrt{2})\ =168+56\sqrt{2}$ $=\alpha +\beta \sqrt{2}\ ∴\alpha +\beta =168+56=224$

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

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