Question ID: #563
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 $\alpha+\beta\sqrt{2}$, $\alpha, \beta \in Z$, then $\alpha+\beta$ is equal to:
- (1) 280
- (2) 168
- (3) 210
- (4) 224
Solution:
$$ A = \begin{pmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{pmatrix} $$
$$ A^2 = A \cdot A = \begin{pmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{pmatrix} \begin{pmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{pmatrix} $$
Elements of the $3^{rd}$ row of $A^2$:
$$ R_3 = \left[ 4(2)+4\sqrt{2}(2\sqrt{2})+8(4), \quad 4(2\sqrt{2})+4\sqrt{2}(4)+8(4\sqrt{2}), \quad 4(4)+4\sqrt{2}(4\sqrt{2})+8(8) \right] $$
$$ R_3 = \left[ 8+16+32, \quad 8\sqrt{2}+16\sqrt{2}+32\sqrt{2}, \quad 16+32+64 \right] $$
$$ R_3 = \left[ 56, \quad 56\sqrt{2}, \quad 112 \right] $$
Sum of elements of the $3^{rd}$ row:
$$ S = 56 + 56\sqrt{2} + 112 = 168 + 56\sqrt{2} $$
Given $S = \alpha + \beta\sqrt{2}$, we get $\alpha = 168$ and $\beta = 56$.
$$ \alpha + \beta = 168 + 56 = 224 $$
Ans. (4)
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