Let A = [aij] be a matrix of order 3 × 3, with aij=(2)i+j. If the sum of all the elements in thethird row of A2 is α+β2,α,β∈Z, then α+β is equal to

A=(2)2(2)3(2)4(2)3(2)4(2)5(2)4(2)5(2)6A=2224224424428A2=221222222222412222222224 =4−−−−−−(2+4+8)(22+42+82)(4+8+16)sum of elements of 3rd row = 4(14+142+28)=4(42+142)=168+562α+β2∴α+β=168+56=224

Let A = [aij] be a matrix of order 3 × 3, with aij=(2)i+j. If the sum of all the elements in thethird row of A2 is α+β2,α,β∈Z, then α+β is equal to

A=(2)2(2)3(2)4(2)3(2)4(2)5(2)4(2)5(2)6A=2224224424428A2=221222222222412222222224 =4−−−−−−(2+4+8)(22+42+82)(4+8+16)sum of elements of 3rd row = 4(14+142+28)=4(42+142)=168+562α+β2∴α+β=168+56=224

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

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280

168

210

224

answer is D.

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Detailed Solution

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