Question bank/JEE Main/Mathematics
JEE MainMathematicsToughMatrix
If $A = \begin{bmatrix}
cos\theta & - sin\theta \\
sin\theta & cos\theta
\end{bmatrix}$, then the
matrix is equal to
Show answer and solution
$A = \begin{bmatrix}
\cos\theta & - \sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},{adj}A = \begin{bmatrix}
\cos\theta & \sin\theta \\
- \sin\theta & \cos\theta
\end{bmatrix}\RightarrowA^{- 1} = \begin{bmatrix}
\cos\theta & \sin\theta \\
- \sin\theta & \cos\theta
\end{bmatrix}$
$\left( A^{- 1} \right)^{20} = \begin{bmatrix}
cos2\theta & sin2\theta \\
- sin2\theta & cos2\theta
\end{bmatrix}$
$A^{- 50} = \begin{bmatrix}
cos( - 50\theta) & - sin( - 50\theta) \\
sin( - 50\theta) & cos( - 50\theta)
\end{bmatrix}$
$A^{- 50}at\ \theta = \frac{\pi}{12}\ is\ \begin{bmatrix}
cos\frac{25\pi}{6} & sin\frac{25\pi}{6} \\
- sin\frac{25\pi}{6} & cos\frac{25\pi}{6}
\end{bmatrix} = \begin{bmatrix}
cos\frac{\pi}{6} & sin\frac{\pi}{6} \\
- sin\frac{\pi}{6} & cos\frac{\pi}{6}
\end{bmatrix} = \begin{bmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{- 1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}$
\cos\theta & - \sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},{adj}A = \begin{bmatrix}
\cos\theta & \sin\theta \\
- \sin\theta & \cos\theta
\end{bmatrix}\RightarrowA^{- 1} = \begin{bmatrix}
\cos\theta & \sin\theta \\
- \sin\theta & \cos\theta
\end{bmatrix}$
$\left( A^{- 1} \right)^{20} = \begin{bmatrix}
cos2\theta & sin2\theta \\
- sin2\theta & cos2\theta
\end{bmatrix}$
$A^{- 50} = \begin{bmatrix}
cos( - 50\theta) & - sin( - 50\theta) \\
sin( - 50\theta) & cos( - 50\theta)
\end{bmatrix}$
$A^{- 50}at\ \theta = \frac{\pi}{12}\ is\ \begin{bmatrix}
cos\frac{25\pi}{6} & sin\frac{25\pi}{6} \\
- sin\frac{25\pi}{6} & cos\frac{25\pi}{6}
\end{bmatrix} = \begin{bmatrix}
cos\frac{\pi}{6} & sin\frac{\pi}{6} \\
- sin\frac{\pi}{6} & cos\frac{\pi}{6}
\end{bmatrix} = \begin{bmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{- 1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}$
Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx
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