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Inverse of the matrix [(cos 2 theta,-sin...

Inverse of the matrix `[(cos 2 theta,-sin 2theta),(sin 2 theta, cos 2theta)]` is

A

`[(cos 2 theta, -sin 2 theta),(sin 2 theta, cos 2 theta)]`

B

`[(cos 2 theta, sin 2theta),(sin 2 theta, -cos 2 theta)]`

C

`[(cos 2 theta, -sin 2 theta),(sin 2 theta, cos 2 theta)]`

D

`[(cos 2 theta, sin 2 theta),(-sin 2 theta, cos 2 theta)]`

Text Solution

Verified by Experts

The correct Answer is:
D
Here cofactors of given matrix are
`C_(11)=cos 2 theta, C_(12)=-sin 2 theta`
`C_(21)=sin 2 theta, C_(22)=cos 2 theta`
and `|A|=cos^(2)2 theta+sin^(2) 2 theta=1`
`:.A^(-1)=1/(|A|)[(C_(11),C_(12)),(C_(21),C_(22))]^(T)=1/(|1|)[(cos 2 theta, -sin 2 theta),(sin 2 theta, cos 2 theta)]^(T)`
`=[(cos 2 theta, sin 2 theta),(-sin 2 theta, cos 2 theta)]`
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Knowledge Check

  • Inverse of the matrix [{:(cos 2 theta, -sin 2 theta),(sin 2 theta, cos 2 theta):}] is

    A
    `[{:(cos 2 theta, -sin 2 theta),(sin 2 theta, cos 2 theta):}]`
    B
    `[{:(cos 2 theta, sin 2 theta),(sin 2 theta, -cos 2 theta):}]`
    C
    `[{:(cos 2 theta, sin 2 theta),(sin 2 theta, cos 2 theta):}]`
    D
    `[{:(cos 2 theta, sin 2 theta),(-sin 2 theta, cos 2 theta):}]`

  • Inverse of the matrix A=[[cos2theta, -sin2theta], [sin2theta, cos2theta]] is

    A
    `[[cos2theta, sin2theta], [sin2theta, cos2theta]]`
    B
    `[[cos2theta, sin2theta], [-sin2theta, cos2theta]]`
    C
    `[[cos2theta, -sin2theta], [sin2theta, cos2theta]]`
    D
    `[[cos2theta, sin2theta], [sin2theta, -cos2theta]]`

  • (1+sin 2theta+cos 2theta)/(1+sin2 theta-cos 2 theta) =

    A
    `(1)/(2)tan theta`
    B
    `(1)/(2)cot theta`
    C
    `tan theta`
    D
    `cot theta`

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