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Question

Let [an+1bn+1]=[3+113311+3][anbn], nN. If a1=b1=1 and a22=2n. Then the value of n is

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Solution

Let A=[anbn]

=[3+113311+3][an1bn1]

sinπ12=3122, cosπ12=3+122
A=22⎢ ⎢cosπ12sinπ12sinπ12cosπ12⎥ ⎥[an1bn1]

=(22)2⎢ ⎢cosπ12sinπ12sinπ12cosπ12⎥ ⎥2[an2bn2]
...
=(22)n1⎢ ⎢cosπ12sinπ12sinπ12cosπ12⎥ ⎥n1[a1b1]

[a22b22]=(22)21⎢ ⎢cosπ12sinπ12sinπ12cosπ12⎥ ⎥21[11]

We know that,
[cosxsinxsinxcosx]n=[cosnxsinnxsinnxcosnx]

[a22b22]=(22)21⎢ ⎢cos7π4sin7π4sin7π4cos7π4⎥ ⎥[11]

=(22)21⎢ ⎢ ⎢ ⎢12121212⎥ ⎥ ⎥ ⎥[11]

=231[1111][11]

=232[10]

Hence, a22=232n=32

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