# Need some help with Matrix!

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!! Note by Abhineet Nayyar
6 years ago

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## Comments

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Is the answer$(2\sqrt{2})^{21}$ ?

- 6 years ago

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Yes I also getting same ..... what's ur method ?

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

- 6 years ago

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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 $b_n = (2 \sqrt{2} )^{21} \sin 360^{\circ} =0$

- 6 years ago

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

So 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...

- 6 years ago

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Also I like your status very much ..... :)

- 6 years ago

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Haha...Thank You!!:D:D

- 6 years 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:)

- 6 years ago

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

- 6 years ago

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

- 6 years ago

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Thanks ... we are egarly waiting ....

- 6 years ago

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Thanks @Sudeep Salgia and @Karan Shekhawat for your help:)

- 6 years ago

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