For any arbitrary matrix,

\(\left[ \begin{matrix} { a }_{ n+1 } \\ { b }_{ n+1 } \end{matrix} \right] \quad =\quad \begin{bmatrix} \sqrt { 3 } +1 & \quad 1-\sqrt { 3 } \\ \sqrt { 3 } -1 & 1+\sqrt { 3 } \end{bmatrix}\quad \left[ \begin{matrix} { a }_{ n } \\ { b }_{ n } \end{matrix} \right] \)

where \(n\in N\) Also, \({ a }_{ n }={ b }_{ n }=1\)

Find \({ a }_{ 22 }\)

P.S. Thanks to my friend Shantam, for the question!!

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TopNewestIs the answer\( (2\sqrt{2})^{21} \) ? – Sudeep Salgia · 2 years, 1 month ago

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I did in this way......

\(\\ \begin{cases} \vec { { V }_{ n } } ={ a }_{ n }i+{ b }_{ n }j \\ \vec { { V }_{ n+1 } } ={ a }_{ n+1 }i+{ b }_{ n+1 }j \end{cases}\\ \)

Now using matrix manuipaltion method that axis is rotted by an angle of 15 degrre ...

\(\displaystyle{\vec { { V }_{ n+1 } } =(2\sqrt { 2 } )\vec { { V }_{ n } } \\ \vec { { V }_{ 1 } } =i+j\quad (\because \quad a_{ 1 }=b_{ 1 }=1)\\ \\ \vec { { V }_{ 22 } } =(2\sqrt { 2 } )\vec { { V }_{ 21 } } ={ \left( 2\sqrt { 2 } \right) }^{ 21 }\vec { { V }_{ 1 } } }\)

Now simply comparing coffecients of An and Bn we get answer.....

Is it correct ....? @Sudeep Salgia – Karan Shekhawat · 2 years, 1 month ago

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– Sudeep Salgia · 2 years, 1 month ago

Yep. There is a small typo in the first equation. It should be \( \displaystyle \vec{ V_{n+1} } = ( 2 \sqrt{2} ) e^{\frac{i \pi }{12}} \vec{V_n } \) And the net rotation is \( 315^{ \circ } \)with the initial angle being \( 45^{\circ } \) making the angle \( 360^{\circ} \). Hence \( a_n = (2 \sqrt{2} )^{21} \cos 360^{\circ} \) and \( b_n = (2 \sqrt{2} )^{21} \sin 360^{\circ} =0 \)Log in to reply

This in which deepanshu gupta posted solution by using this concept ....!

Yes thanks... I missed in typing ... And Abhineet I don't think there is another method 0 ... if fact This question is designed using the concept of Matrix Transformation method ... which I learnt from here brilliant in a questionSo I think this question is specially designed for this concept... May be possible some other method but I think they will surly not an elegant one (at least time consuming) ... But I don't have any ideas about them yet... – Karan Shekhawat · 2 years, 1 month ago

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@Karan Shekhawat Thanks, actually i found it in an MTG magazine for engineering, well, my friend did...So, it is possible that the question may be from vectors, and not matrices. Anyways, Solutions are welcome. So, let's see:) – Abhineet Nayyar · 2 years, 1 month ago

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– Karan Shekhawat · 2 years, 1 month ago

Also I like your status very much ..... :)Log in to reply

– Abhineet Nayyar · 2 years, 1 month ago

Haha...Thank You!!:D:DLog in to reply

– Abhineet Nayyar · 2 years, 1 month ago

Thanks guys,but is there a method to solve it by Matrix algebra and not Vectors...?Log in to reply

– Sudeep Salgia · 2 years, 1 month ago

I have a method which is not so intuitive and I will post it as soon as I get some time.Log in to reply

@Sudeep Salgia and @Karan Shekhawat for your help:) – Abhineet Nayyar · 2 years, 1 month ago

ThanksLog in to reply

– Karan Shekhawat · 2 years, 1 month ago

Thanks ... we are egarly waiting ....Log in to reply