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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
If $A =\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right]$, then :
  • A
    $A ^{30}- A ^{25}=2 I$
  • B
    $A ^{30}+ A ^{25}+ A = I$
  • ✓
    $A ^{30}+ A ^{25}- A = I$
  • D
    $A ^{30}= A ^{25}$
✓

Answer

Correct option: C.
$A ^{30}+ A ^{25}- A = I$
$ A =\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right] A =\left[\begin{array}{cc}\cos 60^{\circ} & \sin 60^{\circ} \\ -\sin 60^{\circ} & \cos 60^{\circ}\end{array}\right]$
$\text { If } A =\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
Here $\alpha=\frac{\pi}{3} ~ A ^2=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
$\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right] ~$
$ =\left[\begin{array}{cc}\cos 2 \alpha & \sin 2 \alpha \\ -\sin 2 \alpha & \cos 2 \alpha\end{array}\right]$
$A ^{30}=\left[\begin{array}{cc}\cos 30 \alpha & \sin 30 \alpha \\ -\sin 30 \alpha & \cos 30 \alpha\end{array}\right] ~$
$ A ^{30}=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]= I$
$A^{25}=\left[\begin{array}{cc}\cos 25 \alpha & \sin 25 \alpha \\ -\sin 25 \alpha & \cos 25 \alpha\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{-\sqrt{3}}{2} & \frac{1}{2}\end{array}\right]$
$A^{25}=A$
$A^{25}-A=0$

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