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By matrix multiplication from that M=[[c...

By matrix multiplication from that `M=[[cos theta,-sin theta],[sin theta,cos theta]]` is orthogonal.

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determinant |"cos theta sin theta -sin theta cos theta

Evaluate : |(cos theta, sin theta),(-sin theta,cos theta)|

Knowledge Check

  • The inverse of A=[[cos theta, sin theta],[-sin theta,cos theta]] is :

    A
    A
    B
    `-A`
    C
    `A^(T)`
    D
    `-A^(T)`

  • If A=[[cos theta,sin theta],[-sin theta,cos theta]] and A(adj A)=[[k,0],[0,k]] , then k =

    A
    0
    B
    `sin theta`
    C
    `cos theta`
    D
    1

  • If A=[{:(cos theta, sin theta),(-sin theta, cos theta):}] then "A A"^(T) is :

    A
    A
    B
    `A^(T)`
    C
    I
    D
    0

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    Solve sin 2 theta+cos theta=0 .

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    ((cos theta+isin theta)^4)/((sin theta+icos theta)^5)=?

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