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If A=[(cos alpha, sin alpha),(-sin alpha...

If `A=[(cos alpha, sin alpha),(-sin alpha, cos alpha)]`, then verify that `A'A=I`

A

`(pi)/(32)`

B

0

C

`(pi)/(64)`

D

`(pi)/(16)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given, matrix `A = [{:("cos" alpha, -"sin"alpha),("sin" alpha, "cos"alpha):}]`
`therefore A^(2) = [{:("cos" alpha, -"sin"alpha),("sin" alpha, "cos"alpha):}][{:("cos" alpha, -"sin"alpha),("sin" alpha, "cos"alpha):}]`
`= [{:("cos"^(2) alpha -"sin"^(2)alpha, -"cos"alpha"sin"alpha-"sin"alpha"cos"alpha),("sin" alpha"cos"alpha + "cos"alpha"sin"alpha, " sin"^(2)alpha + "cos"^(2)alpha):}]`
`= [{:("cos"2 alpha, -"sin"2alpha),("sin" 2alpha, "cos"2alpha):}]`
Similarly,
`A^(n) = [{:("cos"(n alpha), -"sin"(nalpha)),("sin" (nalpha), "cos"(nalpha)):}], n in N`
`rArr A^(32)= [{:("cos"(32 alpha), -"sin"(32 alpha)),("sin" (32 alpha), "cos"( 32 alpha)):}] = [{:(0, -1), (1, 0):}]` (given)
So, `"cos"(32 alpha) = 0 " and sin"(32 alpha) = 1`
`rArr 32 alpha = (pi)/(2) rArr alpha = (pi)/(64)`
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