\(\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\cos \mathrm{x} & 0 & \sin \mathrm{x} \\ 0 & 1 & 0 \\ -\sin \mathrm{x} & 0 & \cos \mathrm{x}\end{array}\right|, \mathrm{g}(\mathrm{y})=\left|\begin{array}{ccc}\cos \mathrm{y} & -\sin \mathrm{y} & 0 \\ \sin \mathrm{y} & \cos \mathrm{y} & 0 \\ 0 & 0 & 1\end{array}\right|\) (i) \(\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y})=\) (a) \(\mathrm{f}(\mathrm{xy})\) (b) \(\mathrm{f}(\mathrm{x} / \mathrm{y})\) (c) \(\mathrm{f}(\mathrm{x}+\mathrm{y})\) (d) \(\mathrm{f}(\mathrm{x}-\mathrm{y})\) (ii) Which of the following is correct? (a) \([\mathrm{f}(\mathrm{x})]^{-1}=[1 /\\{\mathrm{f}(\mathrm{x})\\}]\) (b) \([\mathrm{f}(\mathrm{x})]^{-1}=-\mathrm{f}(\mathrm{x})\) (c) \([\mathrm{f}(\mathrm{x})]^{-1}=\mathrm{f}(-\mathrm{x})\) (d) \([\mathrm{f}(\mathrm{x})]^{-1}=-\mathrm{f}(-\mathrm{x})\) (iii) \([\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{y})]^{-1}=\) (a) \(\mathrm{f}\left(\mathrm{x}^{-1}\right) \mathrm{g}\left(\mathrm{y}^{-1}\right)\) (b) \(\mathrm{f}\left(\mathrm{y}^{-1}\right) \mathrm{g}\left(\mathrm{x}^{-1}\right)\) (c) \(\mathrm{f}(-\mathrm{x}) \mathrm{g}(-\mathrm{y})\) (d) \(\mathrm{g}(-\mathrm{y}) \mathrm{f}(-\mathrm{x})\)

Step by step solution

01

Compute the product f(x) * f(y)

To compute the product f(x) * f(y), we need to multiply their corresponding matrices: \[ \left(\begin{array}{ccc} \cos x & 0 & \sin x \\ 0 & 1 & 0 \\ -\sin x & 0 & \cos x \end{array}\right) \times \left(\begin{array}{ccc} \cos y & 0 & \sin y \\ 0 & 1 & 0 \\ -\sin y & 0 & \cos y \end{array}\right) \] By multiplying, we get: \[ \left(\begin{array}{ccc} \cos x \cos y-\sin x \sin y & 0 & \cos x \sin y+\sin x \cos y \\ 0 & 1 & 0 \\ -\sin x \cos y-\cos x \sin y & 0 & -\sin x \sin y+\cos x \cos y \end{array}\right) \] We recognize this as the matrix f(x+y), so the correct answer is (c) \(f(x+y)\).

02

Compute the inverse of f(x)

To compute the inverse of the matrix f(x): \[ \left(\begin{array}{ccc} \cos x & 0 & \sin x \\ 0 & 1 & 0 \\ -\sin x & 0 & \cos x \end{array}\right)^{-1} \] Since f(x) represents a rotation matrix in 3 dimensions, the inverse matrix is just its transpose: \[ \left(\begin{array}{ccc} \cos x & 0 & -\sin x \\ 0 & 1 & 0 \\ \sin x & 0 & \cos x \end{array}\right) \] This corresponds to the matrix f(-x), so the correct answer is (c) \(f(-x)\).

03

Compute the inverse of f(x) * g(y)

We want to compute the inverse of the matrix product f(x) * g(y): \[ \left(\begin{array}{ccc} \cos x & 0 & \sin x \\ 0 & 1 & 0 \\ -\sin x & 0 & \cos x \end{array}\right) \times \left(\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right)^{-1} \] Recall that for a product of invertible matrices A and B, the inverse of their product is given by (AB)^{-1} = B^{-1} A^{-1}. Thus, we just have to compute the inverses we already found in the previous steps: \[ \left(g(y)^{-1}\right) \times \left(f(x)^{-1}\right) = \left(\begin{array}{ccc} \cos y & \sin y & 0 \\ -\sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right) \times \left(\begin{array}{ccc} \cos x & 0 & -\sin x \\ 0 & 1 & 0 \\ \sin x & 0 & \cos x \end{array}\right) \] As we can see, this corresponds to \(g(-y) f(-x)\), so the correct answer is (d) \(g(-y) f(-x)\).

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