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JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant

यदि \(A =\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\) हो, तो आव्यूह \(A ^{-50}\) होगा, जब \(\theta=\frac{\pi}{12}\) हो

  1. A \(\left[ {\begin{array}{*{20}{c}}
    {\frac{1}{2}}&{ - \frac{{\sqrt 3 }}{2}}\\
    {\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}
    \end{array}} \right]\)
  2. B \(\left[ {\begin{array}{*{20}{c}}
    {\frac{{\sqrt 3 }}{2}}&{ - \frac{1}{2}}\\
    {\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}
    \end{array}} \right]\)
  3. C \(\left[ {\begin{array}{*{20}{c}}
    {\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
    { - \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}
    \end{array}} \right]\)
  4. D \(\left[ {\begin{array}{*{20}{c}}
    {\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\\
    { - \frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}
    \end{array}} \right]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left[ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
{ - \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}
\end{array}} \right]\)

Step-by-step Solution

Detailed explanation

\(A = \left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }\\ {\sin \theta }&{\cos \theta } \end{array}} \right]\)…
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