Let be a matrix of order , with . If the sum of all the elements in the third row of is , then is equal to : [2025]

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Solution

Q.
Let $A = [a_{i j}]$ 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 $α + β \sqrt{2}, α, β \in Z$ , then $α + β$ is equal to : [2025]

1 168

2 224

3 210

4 280

Ans.
(2)

We have,

$A = [a_{i j}] of order 3 \times 3 with a_{i j} = {(\sqrt{2})}^{i + j}$

$A = [\begin{matrix} {(\sqrt{2})}^{2} & {(\sqrt{2})}^{3} & {(\sqrt{2})}^{4} \\ {(\sqrt{2})}^{3} & {(\sqrt{2})}^{4} & {(\sqrt{2})}^{5} \\ {(\sqrt{2})}^{4} & {(\sqrt{2})}^{5} & {(\sqrt{2})}^{6} \end{matrix}] = [\begin{matrix} 2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8 \end{matrix}] = 2 [\begin{matrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{matrix}]$

$\Rightarrow A^{2} = 4 [\begin{matrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{matrix}] [\begin{matrix} 1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4 \end{matrix}] = 4 [\begin{matrix} 7 & 7 \sqrt{2} & 14 \\ 7 \sqrt{2} & 14 & 14 \sqrt{2} \\ 14 & 14 \sqrt{2} & 28 \end{matrix}]$

$∴$ Sum of elements of third row = $4 (14 + 14 \sqrt{2} + 28)$

$= 4 (42 + 14 \sqrt{2}) = 168 + 56 \sqrt{2}$

Comparing the equation $α + β \sqrt{2}$ , we get

$∴ α = 168, β = 56 \Rightarrow α + β = 224$ .

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