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Let n2 be an integer
A=[cos(2πn)sin(2πn)0sin(2πn)cos(2πn)0001] and I is the identity matrix of order 3, then following of which is correct
A.An=I and An1I
B.AmI for any positive integer m
C.A is not invertible
D.An=0 for a positive integer m

Answer
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Hint: Here we will first find the square of the given matrix and then we will find its cube. Then we will see that it is forming a certain pattern. We will follow the same pattern to find the matrix raised to given power. We will simplify the matrix using trigonometric identities and find the correct answer.

Complete step-by-step answer:
The given matrix is A=[cos(2πn)sin(2πn)0sin(2πn)cos(2πn)0001].
Let 2πn=x and we will substitute this value here.
A=[cosxsinx0sinxcosx0001]
Now, we will find the product of matrices A×A.
We can write A×A as A2.
A×A=A2=[cosxsinx0sinxcosx0001]×[cosxsinx0sinxcosx0001]
Now, we will multiply these matrices using the rule of multiplication of matrices. Therefore, we get
A2=[cosx×cosx+sinx×(sinx)cosx×sinx+sinx×cosx0(sinx)×cosx+cosx×(sinx)(sinx)×sinx+cosx×cosx0001]
On further simplifying the terms, we get
A2=[cos2xsin2x2cosxsinx02cosxsinxcos2xsin2x0001]
Now using the trigonometric identities 2sinxcosx=sin2xand cos2xsin2x=cos2x in the above matrix, we get.
A2=[cos2xsin2x0sin2xcos2x0001]
Similarly, we will find the product of matrices A2×A .
We can write A2×A as A3.
A2×A=A3=[cos2xsin2x0sin2xcos2x0001]×[cosxsinx0sinxcosx0001]
Now, we will multiply these matrices using the rule of multiplication of matrices.
A2=[cos2x×cosx+sin2x×(sinx)cos2x×sinx+sin2x×cosx0(sin2x)×cosx+cos2x×(sinx)(sin2x)×sinx+cos2x×cosx0001]

Now using the trigonometric identities sinAcosB+cosAsinB=sin(A+B) and cosAcosBsinAsinB=cos(A+B) in the above matrix, we get
A3=[cos3xsin3x0sin3xcos3x0001]
We can see that it is following a certain pattern as shown below:
An=[cosnxsinnx0sinnxcosnx0001]
Now, we will substitute the value 2πn=x in the above matrix. Therefore, we get
An=[cosn2πnsinn2πn0sinn2πncosn2πn0001]
On further multiplying the terms, we get
An=[cos2πsin2π0sin2πcos2π0001]

Now, substituting sin2π=0 and cos2π=1 in the above matrix, we get
An=[100010001]
We know that I=[100010001]
Therefore, we have
An=[100010001]=IAn=I
Also, An1I
Hence, the correct option is option A.

Note: To solve this question, we need to know the meaning or definition of the trigonometric identities. Trigonometric identities are defined as the equalities which involve the trigonometric functions. They are always true for every value of the occurring variables for which both sides of the equality are defined. We need to remember that all the trigonometric identities are periodic in nature. They repeat their values after a certain interval. These intervals are a multiple of 2π.


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