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Matrices and Determinants question

2019 · 9 Jan · Shift 1 · Q24
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  5. /2019 · 9 Jan · Shift 1 · Q24

Matrices and Determinants question

2019 · 9 Jan · Shift 1 · Q24

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]A = \left[ {\begin{matrix} {\cos \theta } & { - \sin \theta } \\ {\sin \theta } & {\cos \theta } \\ \end{matrix} } \right]A=[cosθsinθ​−sinθcosθ​], then the matrix A–50 when θ\thetaθ=π12\pi \over 1212π​, is equal to :
  1. A
    [32−121232]\left[ {\begin{matrix} { {{\sqrt 3 } \over 2}} & { - {1 \over 2}} \\ {{{ 1} \over 2}} & {{{\sqrt 3 } \over 2}} \\ \end{matrix} } \right][23​​21​​−21​23​​​]
  2. B
    [12−3232−12]\left[ {\begin{matrix} {{1 \over 2}} & -{{{\sqrt 3 } \over 2}} \\ {{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} \\ \end{matrix} } \right][21​23​​​−23​​2−1​​]
  3. C
    [3212−1232]\left[ {\begin{matrix} {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \\ -{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \\ \end{matrix} } \right][23​​−21​​21​23​​​]
  4. D
    [1232−3212]\left[ {\begin{matrix} {{1 \over 2}} & {{{\sqrt 3 } \over 2}} \\ {-{{\sqrt 3 } \over 2}} & {{{ 1} \over 2}} \\ \end{matrix} } \right][21​−23​​​23​​21​​]
View written solutionFree

Correct answer: C

  1. The given matrix is
A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]A=\begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix}A=[cosθsinθ​−sinθcosθ​]

This is the standard rotation matrix. For such a matrix,

An=[cos⁡(nθ)−sin⁡(nθ)sin⁡(nθ)cos⁡(nθ)].A^n=\begin{bmatrix} \cos(n\theta) & -\sin(n\theta)\\ \sin(n\theta) & \cos(n\theta) \end{bmatrix}.An=[cos(nθ)sin(nθ)​−sin(nθ)cos(nθ)​].
  1. We need to find A−50A^{-50}A−50 when θ=π12\theta=\frac{\pi}{12}θ=12π​.

Since

A−1=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]=[cos⁡(−θ)−sin⁡(−θ)sin⁡(−θ)cos⁡(−θ)],A^{-1}=\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} =\begin{bmatrix} \cos(-\theta) & -\sin(-\theta)\\ \sin(-\theta) & \cos(-\theta) \end{bmatrix},A−1=[cosθ−sinθ​sinθcosθ​]=[cos(−θ)sin(−θ)​−sin(−θ)cos(−θ)​],

therefore

A−50=[cos⁡(−50θ)−sin⁡(−50θ)sin⁡(−50θ)cos⁡(−50θ)].A^{-50}=\begin{bmatrix} \cos(-50\theta) & -\sin(-50\theta)\\ \sin(-50\theta) & \cos(-50\theta) \end{bmatrix}.A−50=[cos(−50θ)sin(−50θ)​−sin(−50θ)cos(−50θ)​].
  1. Substitute θ=π12\theta=\frac{\pi}{12}θ=12π​:
−50θ=−50⋅π12=−25π6.-50\theta=-50\cdot \frac{\pi}{12}=-\frac{25\pi}{6}.−50θ=−50⋅12π​=−625π​.

Reduce modulo 2π2\pi2π:

−25π6+4π=−25π6+24π6=−π6.-\frac{25\pi}{6}+4\pi=-\frac{25\pi}{6}+\frac{24\pi}{6}=-\frac{\pi}{6}.−625π​+4π=−625π​+624π​=−6π​.

So,

A−50=[cos⁡(−π6)−sin⁡(−π6)sin⁡(−π6)cos⁡(−π6)].A^{-50}=\begin{bmatrix} \cos\left(-\frac{\pi}{6}\right) & -\sin\left(-\frac{\pi}{6}\right)\\ \sin\left(-\frac{\pi}{6}\right) & \cos\left(-\frac{\pi}{6}\right) \end{bmatrix}.A−50=[cos(−6π​)sin(−6π​)​−sin(−6π​)cos(−6π​)​].
  1. Use trigonometric values:
cos⁡(−π6)=cos⁡π6=32,sin⁡(−π6)=−12.\cos\left(-\frac{\pi}{6}\right)=\cos\frac{\pi}{6}=\frac{\sqrt3}{2}, \qquad \sin\left(-\frac{\pi}{6}\right)=-\frac12.cos(−6π​)=cos6π​=23​​,sin(−6π​)=−21​.

Thus,

A−50=[3212−1232].A^{-50}=\begin{bmatrix} \frac{\sqrt3}{2} & \frac12\\ -\frac12 & \frac{\sqrt3}{2} \end{bmatrix}.A−50=[23​​−21​​21​23​​​].
  1. Comparing with the given options, this matches Option C.
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