If A=[cosθ−sinθsinθcosθ], then the matrix A−50 when θ=π12, is equal to :
A
⎡⎢⎣√3212−12√32⎤⎥⎦
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B
⎡⎢⎣√32−1212√32⎤⎥⎦
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C
⎡⎢⎣12√32−√3212⎤⎥⎦
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D
⎡⎢⎣12−√32√3212⎤⎥⎦
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Solution
The correct option is A⎡⎢⎣√3212−12√32⎤⎥⎦ Given, the matrix A=[cosθ−sinθsinθcosθ] AT=[cosθsinθ−sinθcosθ] ∴AAT=I so A is a orthogonal matrix. ⇒A−1=AT⇒A−50=(A50)T Now finding A2 and then A3, A2=[cos2θ−sin2θsin2θcos2θ]
A3=A2.A=[cos3θ−sin3θsin3θcos3θ]
∴An=[cosnθ−sinnθsinnθcosnθ]....(i)
Now evaluating 50θ=50.π12=25π6=4π+π6
Using equation (i), ∴A50=[cos50θ−sin50θsin50θcos50θ]