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If A = [{:("cos" theta, -"sin"theta),("s...

If `A = [{:("cos" theta, -"sin"theta),("sin"theta, "cos"theta):}]`, then the matrix `A^(-50) " when " theta = (pi)/(12)`, is equal to

A

`[{:(" "(1)/(2), (sqrt(3))/(2)),(-(sqrt(3))/(2), (1)/(2)):}]`

B

`[{:((sqrt(3))/(2), -(1)/(2)),((1)/(2), (sqrt(3))/(2)):}]`

C

`[{:((sqrt(3))/(2), (1)/(2)),(-(1)/(2), (sqrt(3))/(2)):}]`

D

`[{:((1)/(2), (-sqrt(3))/(2)),((sqrt(3))/(2), " "(1)/(2)):}]`

Text Solution

Verified by Experts

We have, `A =|{:("cos"theta, -"sin"theta),("sin"theta,"cos"theta):}|`
`therefore |A| = "cos"^(2) theta + "sin"^(2)theta = 1`
and adj `A =|{:("cos"theta, "sin"theta),(-"sin"theta,"cos"theta):}|`
`[because "If" A = [{:(a, b),(c, d):}], " then adj "A = [{:(d, -b),(-c, a):}]]`
`rArr A^(-1) = [{:("cos"theta, "sin"theta),(-"sin"theta,"cos"theta):}] " " (because A^(-1) = ("adj"A)/(|A|))`
Note that, `A^(-50) = (A^(-1))^(50)`
Now, `A^(-2) = (A^(-1))(A^(-1))`
`rArr A^(-2) = [{:("cos"theta, "sin"theta),(-"sin"theta,"cos"theta):}][{:("cos"theta, "sin"theta),(-"sin"theta,"cos"theta):}]`
`= [{:(" cos"^(2)theta-"sin"^(2)theta, "cos"theta"sin"theta + "sin"theta"cos"theta),(-"cos"theta"sin"theta-"cos"theta"sin"theta,-"sin"^(2)theta +"cos"^(2)theta):}]`
`=[{:("cos"2theta, "sin"2theta),(-"sin"2theta,"cos"2theta):}]`
Also, `A^(-3) = (A^(-2))(A^(-1))`
`A^(-3)=[{:("cos"2theta, "sin"2theta),(-"sin"2theta,"cos"2theta):}][{:("cos"theta, "sin"theta),(-"sin"theta,"cos"theta):}]`
`=[{:("cos"3theta, "sin"3theta),(-"sin"3theta,"cos"3theta):}]`
Similarly, `A^(-50) = =[{:("cos"50theta, "sin"50theta),(-"sin"50theta,"cos"50theta):}]`
`=[{:("cos"(25)/(6)pi, "sin"(25)/(6)pi),(-"sin"(25)/(6)pi,"cos"(25)/(6)pi):}] " " ("when" theta = (pi)/(12))`
`=[{:("cos"(pi)/(6), "sin"(pi)/(6)),(-"sin"(pi)/(6),"cos"(pi)/(6)):}] [{:(because "cos"((25pi)/(6)) ="cos"(4pi + (pi)/(6)) ="cos"(pi)/(6)),("and sin"((25pi)/(6)) = "sin"(4pi + (pi)/(6)) = "sin" (pi)/(6)):}]`
`=[{:((sqrt(3))/(2),(1)/(2)),((-1)/(2), (sqrt(3))/(2)):}]`
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