Consider the matrix f ( x ) = [ cos x - sin x 0 sin x cos x 0 0 0 1 ] . Given below are two statements: Statement I : f ( - x ) is the inverse of the matrix f ( x ) . Statement II : f ( x ) f ( y ) = f ( x + y ) . [2024]

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Solution

Q.
Consider the matrix $f (x) =$ $[\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}]$ .

Given below are two statements:

Statement I : $f (- x)$ is the inverse of the matrix $f (x) .$

Statement II : $f (x) f (y) = f (x + y) .$

In the light of the above statements, choose the correct answer from the options given below [2024]

1 Statement I is true but Statement II is false

2 Both Statement I and Statement II are false

3 Statement I is false but Statement II is true

4 Both Statement I and Statement II are true

Ans.
(4)

We have, $f (x) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}]$

$f (- x) = [\begin{matrix} \cos x & \sin x & 0 \\ - \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}]$

$∴$    $f (x) f (- x) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} \cos x & \sin x & 0 \\ - \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I \Rightarrow f (- x) = {[f (x)]}^{- 1}$

$∴$   Statement-I is true.

$f (x) f (y) = [\begin{matrix} \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} \cos y & - \sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{matrix}]$

$= [\begin{matrix} \cos x \cos y - \sin x \sin y & \cos x (- \sin y) - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & \sin x (- \sin y) + \cos x \cos y & 0 \\ 0 & 0 & 1 \end{matrix}]$

$= [\begin{matrix} \cos (x + y) & - \sin (x + y) & 0 \\ \sin (x + y) & \cos (x + y) & 0 \\ 0 & 0 & 1 \end{matrix}] = f (x + y)$

$∴$   Statement-II is also true.

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