Question

If A = [ᶜᵒˢₛᵢₙ ⁻ˢᶦⁿ cₒₛ], then the matrix A⁻⁵⁰ when , θ = π/12, is equal to [JEE Main 2019]

Answer 1

If A = [ᶜᵒˢₛᵢₙ ⁻ˢᶦⁿ cₒₛ], then the matrix A⁻⁵⁰ when , θ = π/12, is equal to $\left[\begin{array}{cc}\sqrt{3}/2&1/2& \\-1/2&\sqrt{3} /2\end{array}\right]$

Inverse of a matrix can be found out by,

• $A^{-1}=\frac{1}{\left | A \right |}\cdot adjA$

Its given that

• A =   $\begin{bmatrix} \cos \left ( \theta \right ) & -\sin \left ( \theta \right )\\ \sin \left ( \theta \right )& \cos \left ( \theta \right ) \end{bmatrix}$

• $A^{-1}=\frac{1}{\left | A \right |}adjA=\frac{1}{1}\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$

• $\left ( A^{-1} \right )^{2}=\begin{bmatrix} \cos\theta & \sin \theta \\ - \sin \theta & \cos\theta \end{bmatrix}\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ = =  >

• $A^{-2} = \begin{bmatrix} \cos \left ( 2\theta \right ) &\sin \left ( 2\theta \right ) \\ -\sin \left ( 2\theta \right ) & \cos \left ( 2\theta \right ) \end{bmatrix}$ = = >

• $\left ( A^{-1} \right )^{3}=\begin{bmatrix} \cos \left (3 \theta \right ) & \sin \left (3 \theta \right )\\ - \sin \left ( 3\theta \right )&\cos \left ( 3\theta \right ) \end{bmatrix}$ = =>

By using principle of mathematical induction this can be used for nθ

For n = 50,

• $\left ( A^{-1} \right )^{50}=\begin{bmatrix} \cos \left (50 \theta \right ) & \sin \left (50 \theta \right )\\ - \sin \left ( 50\theta \right )&\cos \left ( 50\theta \right ) \end{bmatrix}$

Given θ = π/12

cos(50$\pi$/12) = cos( 4$\pi$ + $\pi$/6) = cos($\pi$/6) = $\sqrt{3}$/2

sin(50π/12) = sin ( π/6 ) = 1/2

Therefore,

• $A^{-50}$ = $\left[\begin{array}{cc}\sqrt{3}/2&1/2& \\-1/2&\sqrt{3} /2\end{array}\right]$  
• Option C is the right answer

Answer 2

If A = [ᶜᵒˢₛᵢₙ ⁻ˢᶦⁿ cₒₛ], then the matrix A⁻⁵⁰ when , θ = π/12, is equal to \begin{lgathered}\left[\begin{array}{cc}\sqrt{3}/2&1/2& \\-1/2&\sqrt{3} /2\end{array}\right]\end{lgathered}

[

3

/2

−1/2

1/2

3

/2

]

Inverse of a matrix can be found out by,

A^{-1}=\frac{1}{\left | A \right |}\cdot adjAA

−1

=

∣A∣

1

⋅adjA

Its given that

A = \begin{lgathered}\begin{bmatrix} \cos \left ( \theta \right ) & -\sin \left ( \theta \right )\\ \sin \left ( \theta \right )& \cos \left ( \theta \right ) \end{bmatrix}\end{lgathered}

[

cos(θ)

sin(θ)

−sin(θ)

cos(θ)

]

\begin{lgathered}A^{-1}=\frac{1}{\left | A \right |}adjA=\frac{1}{1}\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\end{lgathered}

A

−1

=

∣A∣

1

adjA=

1

1

[

cosθ

−sinθ

sinθ

cosθ

]

\begin{lgathered}\left ( A^{-1} \right )^{2}=\begin{bmatrix} \cos\theta & \sin \theta \\ - \sin \theta & \cos\theta \end{bmatrix}\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\end{lgathered}

(A

−1

)

2

=[

cosθ

−sinθ

sinθ

cosθ

][

cosθ

−sinθ

sinθ

cosθ

]

= = >

\begin{lgathered}A^{-2} = \begin{bmatrix} \cos \left ( 2\theta \right ) &\sin \left ( 2\theta \right ) \\ -\sin \left ( 2\theta \right ) & \cos \left ( 2\theta \right ) \end{bmatrix}\end{lgathered}

A

−2

=[

cos(2θ)

−sin(2θ)

sin(2θ)

cos(2θ)

]

= = >

\begin{lgathered}\left ( A^{-1} \right )^{3}=\begin{bmatrix} \cos \left (3 \theta \right ) & \sin \left (3 \theta \right )\\ - \sin \left ( 3\theta \right )&\cos \left ( 3\theta \right ) \end{bmatrix}\end{lgathered}

(A

−1

)

3

=[

cos(3θ)

−sin(3θ)

sin(3θ)

cos(3θ)

]

= =>

By using principle of mathematical induction this can be used for nθ

For n = 50,

\begin{lgathered}\left ( A^{-1} \right )^{50}=\begin{bmatrix} \cos \left (50 \theta \right ) & \sin \left (50 \theta \right )\\ - \sin \left ( 50\theta \right )&\cos \left ( 50\theta \right ) \end{bmatrix}\end{lgathered}

(A

−1

)

50

=[

cos(50θ)

−sin(50θ)

sin(50θ)

cos(50θ)

]

Given θ = π/12

cos(50\piπ /12) = cos( 4\piπ + \piπ /6) = cos(\piπ /6) = \sqrt{3}

3

/2

sin(50π/12) = sin ( π/6 ) = 1/2

Therefore,

A^{-50}A

−50

= \begin{lgathered}\left[\begin{array}{cc}\sqrt{3}/2&1/2& \\-1/2&\sqrt{3} /2\end{array}\right]\end{lgathered}

[

3

/2

−1/2

1/2

3

/2

]

Option C is the right answer

Hope this helps you