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The multiplicative inverse of A = [(cos ...

The multiplicative inverse of `A = [(cos theta,-sin theta),(sin theta,cos theta)]`is

A

`[{:(-cos theta, sin theta),(-sin theta, -cos theta):}]`

B

`[{:(cos theta, sin theta),(-sin theta, cos theta):}]`

C

`[{:(-cos theta, -sin theta),(sin theta, -cos theta):}]`

D

`[{:(cos theta, sin theta),(sin theta, -cos theta):}]`

Text Solution

Verified by Experts

The correct Answer is:
B
`|A|=cos ^(2)+sin^(2)theta=1`
`adj(A)=[{:(cos theta,sintheta),(-sin theta, cos theta):}]`
`A^(-1)=(adj(A))/(|A|)=[{:(cos theta, sin theta),(-sin theta, cos theta):}]`
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(i) |(cos theta,-sin theta),(sin theta,cos theta)|

A = [(cos theta,-sin theta),(-sin theta,-cos theta)] then find A^(-1) .

Knowledge Check

  • 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)]`

  • Let A=[(cos theta, -sin theta),(- sin theta,-cos theta)] then the inverse of a is

    A
    `[(cos theta, - sin theta),(- sin theta, - cos theta)]`
    B
    `[(-cos theta, sin theta),(sin theta, cos theta)]`
    C
    `[(sin theta, -cos theta),(cos theta, - sin theta)]`
    D
    `[(-sin theta, - cos theta),(- cos theta, sin theta)]`

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