SATHEE: Matrices and Determinants 3 Question 3

Matrices and Determinants 3 Question 3

4. If $A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ , then the matrix $A^{- 50}$ when $θ = \frac{π}{12}$ , is equal to

(2019 Main, 9 Jan I)

(a) $[\begin{matrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ - \frac{\sqrt{3}}{2} & \frac{1}{2} \end{matrix}]$

(b) $[\begin{matrix} \frac{\sqrt{3}}{2} & - \frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}]$

(d) $[\begin{matrix} \frac{1}{2} & - \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{matrix}]$

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Answer:

Correct Answer: 4. (c)

Solution:

We have, $A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$

$∴ | A | = \cos^{2} θ + \sin^{2} θ = 1$

and $adj A = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$

$∵ .$ If $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , then $adj A = [\begin{matrix} d & - b \\ - c & a \end{matrix}]$

$\Rightarrow A^{- 1} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] ∵ A^{- 1} = \frac{adj A}{| A |}$

Note that, $A^{- 50} = {(A^{- 1})}^{50}$

Now, $A^{- 2} = (A^{- 1}) (A^{- 1})$

$\Rightarrow A^{- 2} = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$ $[\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$

= $[\begin{matrix} \cos^{2} θ - \sin^{2} θ & \cos θ \sin θ + \sin θ \cos θ \\ - \cos θ \sin θ - \cos θ \sin θ & - \sin^{2} θ + \cos^{2} θ \end{matrix}]$

= $[\begin{matrix} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{matrix}]$

Also, $A^{- 3} = (A^{- 2}) (A^{- 1})$

$\begin{aligned} A^{- 3} = [\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ - \sin 2 θ & \cos 2 θ \end{array}] [\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] \\ = [\begin{array}{cc} \cos 3 θ & \sin 3 θ \\ - \sin 3 θ & \cos 3 θ \end{array}] \\ Similarly, A^{- 50} = [\begin{array}{cc} \cos 50 θ & \sin 50 θ \\ - \sin 50 θ & \cos 50 θ \end{array}] \\ = [\begin{array}{cc} \cos \frac{25}{6} π & \sin \frac{25}{6} π \\ - \sin \frac{25}{6} π & \cos \frac{25}{6} π \end{array}] \\ (when θ = \frac{π}{12}) \\ = [\begin{array}{cc} \cos \frac{π}{6} & \sin \frac{π}{6} \\ - \sin \frac{π}{6} & \cos \frac{π}{6} \end{array}] [\begin{matrix} ∵ \cos (\frac{25 π}{6}) = \cos (4 π + \frac{π}{6}) = \cos \frac{π}{6} \\ and \sin (\frac{25 π}{6}) = \sin (4 π + \frac{π}{6}) = \sin \frac{π}{6} \end{matrix}] \\ = [\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{array}] \end{aligned}$

Matrices and Determinants 3 Question 3

4. If $A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ , then the matrix $A^{- 50}$ when $θ = \frac{π}{12}$ , is equal to

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Matrices and Determinants 3 Question 3

4. If A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ], then the matrix A−50 when θ=π12, is equal to

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4. If $A = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$ , then the matrix $A^{- 50}$ when $θ = \frac{π}{12}$ , is equal to