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Flind the product of two matrices A =[...

Flind the product of two matrices
`A =[[cos^(2) theta , cos theta sin theta],[cos theta sin theta ,sin^(2)theta]] B= [[cos^(2) phi,cos phi sin phi],[cos phisin phi,sin^(2)phi]]`
Show that, AB is the zero matrix if `theta and phi` differ by an
odd multipl of `pi/2`.

A

5(6!)

B

3(6!)

C

12(6!)

D

8(6!)

Text Solution

Verified by Experts

The correct Answer is:
D
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If A=[(cos^2theta,costheta sintheta),(cos theta sintheta, sin^2theta)] and B=[(cos^2 phi, cos phi sin phi),(cos phisinphi,sin^2phi)], show that AB is zero matrix if theta and phi differ by an odd multiple of pi/2.

If AB=O for the matrices A=[[cos^2theta,costhetasintheta],[costhetasintheta,sin^2theta]] and B=[[cos^2phi,cosphisinphi],[cosphisinphi,sin^2phi]] then theta-phi is

Knowledge Check

  • If AB =0, then for the matrices. A = [{:(cos^(2)theta, cos theta sin theta),(cos theta sin theta, sin^(2)theta):}] and B = [{:(cos^(2)phi, cos phi sin phi),(cos phi sin phi, sin^(2)phi):}], theta - phi is:

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    B
    an odd multiple of `pi`
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    D
    0

  • If A=[(cos^(2) theta, cos theta sin theta),(cos theta sin theta, sin^(2)theta)] and B=[(cos^(2) phi, cos phi sin phi),(cos phi sin phi, sin^(2)phi)] are the two matrices such that the product AB is the null matrix then theta-phi is equal to

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    0
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    C
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  • If AB=0 where A = [(cos^2theta, costhetasintheta),(costhetasintheta,sin^2theta)] and B=[(cos^2phi, cosphisinphi),(cosphisinphi,sin^2phi)] , then |theta-phi| is equal to

    A
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    C
    `pi/2`
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    If A = [[cos^2theta, costhetasintheta],[costhetasintheta, sin^2theta]] B= [[cos^2phi, cosphisinphi], [cosphisinphi, sin^2phi]] and theta - phi = (2n+1)(pi)/2 Find AB.

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