MCQ
If for the matrix, $A =\left[\begin{array}{cc}1 & -\alpha \\ \alpha & \beta\end{array}\right], AA ^{ T }= I _{2},$ then the value of $\alpha^{4}+\beta^{4}$ is ....... .
- A$4$
- B$2$
- C$3$
- ✓$1$
$\Rightarrow\left[\begin{array}{cc}1 & -\alpha \\ \alpha & \beta\end{array}\right]\left[\begin{array}{cc}1 & \alpha \\ -\alpha & \beta\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}1+\alpha^{2} & \alpha-\alpha \beta \\ \alpha-\alpha \beta & \alpha^{2}+\beta^{2}\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow \alpha^{2}=0 \;and\; \beta^{2}=1$
$\therefore \alpha^{4}+\beta^{4}=1$
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$( S 1):|(\overrightarrow{ a } \times \overrightarrow{ b })+(\overrightarrow{ c } \times \overrightarrow{ b })|-|\overrightarrow{ c }|=6(2 \sqrt{2}-1)$
$( S 2): \angle ABC =\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$. Then