Let A=[a_{ij}] be a matrix of order 3 times 3 | vedclass

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(A) Given $A = [a_{ij}]$ where $a_{ij} = (\sqrt{2})^{i+j}$.$A = \begin{bmatrix} (\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \ (\sqrt{2})^3 & (\sqrt{2}

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$224$

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(A) Given $A = [a_{ij}]$ where $a_{ij} = (\sqrt{2})^{i+j}$.$A = \begin{bmatrix} (\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6 \end{bmatrix} = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \ 2\sqrt{2} & 4 & 4\sqrt{2} \ 4 & 4\sqrt{2} & 8 \end{bmatrix}$.We can write $A = 2 \begin{bmatrix} 1 & \sqrt{2} & 2 \ \sqrt{2} & 2 & 2\sqrt{2} \ 2 & 2\sqrt{2} & 4 \end{bmatrix}$.Let $B = \begin{bmatrix} 1 & \sqrt{2} & 2 \ \sqrt{2} & 2 & 2\sqrt{2} \ 2 & 2\sqrt{2} & 4 \end{bmatrix}$,then $A = 2B$.$A^2 = (2B)(2B) = 4B^2$.The third row of $B^2$ is obtained by multiplying the third row of $B$ with the columns of $B$:$R_3(B^2) = [ (2)(1)+(2\sqrt{2})(\sqrt{2})+(4)(2), \quad (2)(\sqrt{2})+(2\sqrt{2})(2)+(4)(2\sqrt{2}), \quad (2)(2)+(2\sqrt{2})(2\sqrt{2})+(4)(4) ]$.$R_3(B^2) = [ 2+4+8, \quad 2\sqrt{2}+4\sqrt{2}+8\sqrt{2}, \quad 4+8+16 ] = [ 14, \quad 14\sqrt{2}, \quad 28 ]$.Sum of elements of the third row of $B^2 = 14 + 14\sqrt{2} + 28 = 42 + 14\sqrt{2}$.Since $A^2 = 4B^2$,the sum of elements of the third row of $A^2 = 4(42 + 14\sqrt{2}) = 168 + 56\sqrt{2}$.Given the sum is $\alpha + \beta\sqrt{2}$,we have $\alpha = 168$ and $\beta = 56$.Therefore,$\alpha + \beta = 168 + 56 = 224$.

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

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