Inverse of the matrix - cos 2- sin 2- -sin 2- cos 2- is

The correct option is D [ cos 2θ sin 2θsin 2θ cos 2θ] Given matrix is cos2θ #160; amp; sin2θ #160; sin2θ #160; amp; cos2θ ...

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Standard XII

Mathematics

Adjoint of a Matrix

Inverse of th...

Question

Inverse of the matrix $[\begin{matrix} cos 2 θ sin 2 θ \end{matrix} \begin{matrix} - sin 2 θ cos 2 θ \end{matrix}]$ is

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

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

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[\begin{matrix} cos 2 θ sin 2 θ \end{matrix} \begin{matrix} sin 2 θ cos 2 θ \end{matrix}]

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

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Solution

The correct option is D $[\begin{matrix} cos 2 θ - sin 2 θ \end{matrix} \begin{matrix} sin 2 θ cos 2 θ \end{matrix}]$
Given matrix is $∣ ∣ ∣ \begin{matrix} cos 2 θ & - sin 2 θ sin 2 θ & cos 2 θ \end{matrix} ∣ ∣ ∣$

Inverse of a matrix of the form $[\begin{matrix} a & b c & d \end{matrix}]$ is given by $\frac{1}{a d - b c} [\begin{matrix} d & - b - c & a \end{matrix}]$

So inverse of the above matrix will be $\frac{1}{{cos}^{2} (2 θ) + {sin}^{2} (2 θ)} [\begin{matrix} cos 2 θ & sin 2 θ - sin 2 θ & cos 2 θ \end{matrix}]$
$\Rightarrow [\begin{matrix} cos 2 θ & sin 2 θ - sin 2 θ & cos 2 θ \end{matrix}]$

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Similar questions

$(\frac{1}{s e c^{2} θ - c o s^{2} θ} + \frac{1}{c o s e c^{2} θ - s i n^{2} θ}) s i n^{2} θ c o s^{2} θ = \frac{1 - s i n^{2} θ c o s^{2} θ}{2 + s i n^{2} θ c o s^{2} θ}$

Q. If A =

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

, find A².

Q. Prove the following trigonometric identities.

(\frac{1}{\sec^{2} θ - \cos^{2} θ} + \frac{1}{{cosec}^{2} θ - \sin^{2} θ}) \sin^{2} θ \cos^{2} θ = \frac{1 - \sin^{2} θ \cos^{2} θ}{2 + \sin^{2} θ \cos^{2} θ}

Q. Prove the following identities (1-17)

(\frac{1}{\sec^{2} θ - \cos^{2} θ} + \frac{1}{{cosec}^{2} θ - \sin^{2} θ}) \sin^{2} θ \cos^{2} θ = \frac{1 - \sin^{2} θ \cos^{2} θ}{2 + \sin^{2} θ \cos^{2} θ}

Q. If

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

and

A + A^{T} = I

, where

I

is the unit of matrix of

2 \times 2

and

A^{T}

is the transpose of

A

, then the value of

θ

is equal to

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Inverse of the matrix [cos2θsin2θ−sin2θcos2θ] is

Inverse of the matrix $[\begin{matrix} cos 2 θ sin 2 θ \end{matrix} \begin{matrix} - sin 2 θ cos 2 θ \end{matrix}]$ is