- Home
- Standard 12
- Mathematics
3 and 4 .Determinants and Matrices
hard
ધારો કે $A =\left( a _{i j}\right)$ એ કક્ષા $3 \times 3$ નો એક શ્રેણિક છે, જ્યાં $a _{i j}=(\sqrt{2})^{i+j}$ છે. જો $A ^2$ ની ત્રીજી હારના તમામ ઘટકોનો સરવાળો $\alpha+\beta \sqrt{2}, \alpha+\beta \in Z$ હોય તો $\alpha+\beta=$ _______.
A$224$
B$168$
C$210$
D$280$
(JEE MAIN-2025)
Solution
$A=\left[\begin{array}{lll}(\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6\end{array}\right]$
$A=\left[\begin{array}{ccc}2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8\end{array}\right]$
$A^2=2^2\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]$
$=4\left[\begin{array}{ccc}- & – & – \\ – & – & – \\ (2+4+8) & (2 \sqrt{2}+4 \sqrt{2}+8 \sqrt{2}) & (4+8+16)\end{array}\right]$
Sum of elements of $3^{\text {rd }}$ row $=4(14+14 \sqrt{2}+28)$
$=4(42+14 \sqrt{2})$
$ =168+56 \sqrt{2}$
$ \alpha+\beta \sqrt{2}$
$\therefore \alpha \alpha+\beta=168+56=224$
$A=\left[\begin{array}{ccc}2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8\end{array}\right]$
$A^2=2^2\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]$
$=4\left[\begin{array}{ccc}- & – & – \\ – & – & – \\ (2+4+8) & (2 \sqrt{2}+4 \sqrt{2}+8 \sqrt{2}) & (4+8+16)\end{array}\right]$
Sum of elements of $3^{\text {rd }}$ row $=4(14+14 \sqrt{2}+28)$
$=4(42+14 \sqrt{2})$
$ =168+56 \sqrt{2}$
$ \alpha+\beta \sqrt{2}$
$\therefore \alpha \alpha+\beta=168+56=224$
Standard 12
Mathematics
