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!!

#Matrices

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Sudeep Salgia - 6 years, 2 months ago

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

Karan Shekhawat - 6 years, 2 months 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 $b_n = (2 \sqrt{2} )^{21} \sin 360^{\circ} =0$

Sudeep Salgia - 6 years, 2 months ago

@Sudeep Salgia – 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...

Karan Shekhawat - 6 years, 2 months ago

@Karan Shekhawat – Also I like your status very much ..... :)

Karan Shekhawat - 6 years, 2 months ago

@Karan Shekhawat – Haha...Thank You!!:D:D

A Former Brilliant Member - 6 years, 2 months ago

@Karan Shekhawat – @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:)

A Former Brilliant Member - 6 years, 2 months ago

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

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – I have a method which is not so intuitive and I will post it as soon as I get some time.

Sudeep Salgia - 6 years, 2 months ago

@Sudeep Salgia – Thanks ... we are egarly waiting ....

Karan Shekhawat - 6 years, 2 months ago

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

A Former Brilliant Member - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$