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Sin(A+B)Sin(A-B)

Let n≥ 2 be...

Question

Let $n \geq 2$ be an integer,

$A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (\frac{2 π}{n}) & sin (\frac{2 π}{n}) & 0 - sin (\frac{2 π}{n}) & cos (\frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$ and $I$ is the identity matrix of order $3$ ., then following of which is correct

A

A^{n} = I

and

A^{n - 1} \neq I

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B

A^{m} \neq I

for any positive integer

m

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C

A

is not invertible

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D

A^{m} = 0

for a positive integer

m

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Solution

The correct option is A $A^{n} = I$ and $A^{n - 1} \neq I$
Given,
$A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (\frac{2 π}{n}) & sin (\frac{2 π}{n}) & 0 - sin (\frac{2 π}{n}) & cos (\frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

Now,
$A \times A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (\frac{2 π}{n}) & sin (\frac{2 π}{n}) & 0 - sin (\frac{2 π}{n}) & cos (\frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \times ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (\frac{2 π}{n}) & sin (\frac{2 π}{n}) & 0 - sin (\frac{2 π}{n}) & cos (\frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {cos}^{2} (\frac{2 π}{n}) - {sin}^{2} (\frac{2 π}{n}) & cos (\frac{2 π}{n}) sin (\frac{2 π}{n}) + sin (\frac{2 π}{n}) cos (\frac{2 π}{n}) & 0 - sin (\frac{2 π}{n}) cos (\frac{2 π}{n}) - sin (\frac{2 π}{n}) cos (\frac{2 π}{n}) & - {sin}^{2} (\frac{2 π}{n}) + {cos}^{2} (\frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (2 \times \frac{2 π}{n}) & 2 sin (\frac{2 π}{n}) cos (\frac{2 π}{n}) & 0 - 2 sin (\frac{2 π}{n}) cos (\frac{2 π}{n}) & cos (2 \times \frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (2 \times \frac{2 π}{n}) & sin (2 \times \frac{2 π}{n}) & 0 - sin (2 \times \frac{2 π}{n}) & cos (2 \times \frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

Similarly,

$A^{n} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (2^{n - 1} \times \frac{2 π}{n}) & sin (2^{n - 1} \times \frac{2 π}{n}) & 0 - sin (2^{n - 1} \times \frac{2 π}{n}) & cos (2^{n - 1} \times \frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = I$

and

$A^{n - 1} = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (2^{n - 2} \times \frac{2 π}{n}) & sin (2^{n - 2} \times \frac{2 π}{n}) & 0 - sin (2^{n - 2} \times \frac{2 π}{n}) & cos (2^{n - 2} \times \frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$\neq I$

$∴ A^{n - 1} \neq I$ for any positive integer $n$ .

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A = ⎛ ⎜ ⎝ \begin{matrix} cos (2 π / n) & sin (2 π / n) & 0 sin (2 π / n) & cos (2 π / n) & 0 0 & 0 & 1 \end{matrix} ⎞ ⎟ ⎠

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