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Equality of Matrices

Let $ beginbm...

Question

Let $[\begin{matrix} a_{n + 1} b_{n + 1} \end{matrix}] = [\begin{matrix} \sqrt{3} + 1 & 1 - \sqrt{3} \sqrt{3} - 1 & 1 + \sqrt{3} \end{matrix}] [\begin{matrix} a_{n} b_{n} \end{matrix}],$ $n \in N .$ If $a_{1} = b_{1} = 1$ and $a_{22} = 2^{n} .$ Then the value of $n$ is

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Solution

Let $A = [\begin{matrix} a_{n} b_{n} \end{matrix}]$

$= [\begin{matrix} \sqrt{3} + 1 & 1 - \sqrt{3} \sqrt{3} - 1 & 1 + \sqrt{3} \end{matrix}] [\begin{matrix} a_{n - 1} b_{n - 1} \end{matrix}]$

$∵ sin \frac{π}{12} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}, cos \frac{π}{12} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$
$A = 2 \sqrt{2} ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{12} & - sin \frac{π}{12} sin \frac{π}{12} & cos \frac{π}{12} \end{matrix} ⎤ ⎥ ⎥ ⎦ [\begin{matrix} a_{n - 1} b_{n - 1} \end{matrix}]$

$= (2 \sqrt{2})^{2} {⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{12} & - sin \frac{π}{12} sin \frac{π}{12} & cos \frac{π}{12} \end{matrix} ⎤ ⎥ ⎥ ⎦}^{2} [\begin{matrix} a_{n - 2} b_{n - 2} \end{matrix}]$
$. . .$
$= (2 \sqrt{2})^{n - 1} {⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{12} & - sin \frac{π}{12} sin \frac{π}{12} & cos \frac{π}{12} \end{matrix} ⎤ ⎥ ⎥ ⎦}^{n - 1} [\begin{matrix} a_{1} b_{1} \end{matrix}]$

$∴ [\begin{matrix} a_{22} b_{22} \end{matrix}] = (2 \sqrt{2})^{21} {⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{π}{12} & - sin \frac{π}{12} sin \frac{π}{12} & cos \frac{π}{12} \end{matrix} ⎤ ⎥ ⎥ ⎦}^{21} [\begin{matrix} 11 \end{matrix}]$

We know that,
${[\begin{matrix} cos x & - sin x sin x & cos x \end{matrix}]}^{n} = [\begin{matrix} cos n x & - sin n x sin n x & cos n x \end{matrix}]$

$∴ [\begin{matrix} a_{22} b_{22} \end{matrix}] = (2 \sqrt{2})^{21} ⎡ ⎢ ⎢ ⎣ \begin{matrix} cos \frac{7 π}{4} & - sin \frac{7 π}{4} sin \frac{7 π}{4} & cos \frac{7 π}{4} \end{matrix} ⎤ ⎥ ⎥ ⎦ [\begin{matrix} 11 \end{matrix}]$

$= (2 \sqrt{2})^{21} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ [\begin{matrix} 11 \end{matrix}]$

$= 2^{31} [\begin{matrix} 1 & 1 - 1 & 1 \end{matrix}] [\begin{matrix} 11 \end{matrix}]$

$= 2^{32} [\begin{matrix} 10 \end{matrix}]$

Hence, $a_{22} = 2^{32} \Rightarrow n = 32$

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