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Mathematics Question on Matrices

Consider the matrix f(x)=[cos⁡x−sin⁡x0 sin⁡xcos⁡x0 001]f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\\ \sin x & \cos x & 0 \\\ 0 & 0 & 1 \end{bmatrix}.
Given below are two statements:
Statement I: f(−x)f(-x) is the inverse of the matrix f(x)f(x).
Statement II: f(x)f(y)=f(x+y)f(x) f(y) = f(x + y).
In the light of the above statements, choose the correct answer from the options given below:"

A

Statement I is false but Statement II is true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Both Statement I and Statement II are true

Answer

Both Statement I and Statement II are true

Explanation

Solution

Step 1. Verification of Statement I: To check if f(−x)f(-x) is the inverse of f(x)f(x), we need to verify if f(x)⋅f(−x)=If(x) \cdot f(-x) = I, where II is the identity matrix.

- Calculate f(−x)f(-x):

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

f(−x)=[cos⁡(−x)−sin⁡(−x)0 sin⁡(−x)cos⁡(−x)0 001]=[cos⁡xsin⁡x0 −sin⁡xcos⁡x0 001]f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \\\ \sin(-x) & \cos(-x) & 0 \\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \\\ -\sin x & \cos x & 0 \\\ 0 & 0 & 1 \end{bmatrix}

- Now, compute f(x)⋅f(−x)f(x) \cdot f(-x):
f(x)⋅f(−x)=[cos⁡x−sin⁡x0 sin⁡xcos⁡x0 001][cos⁡xsin⁡x0 −sin⁡xcos⁡x0 001]=[100 010 001]=If(x) \cdot f(-x) = \begin{bmatrix} \cos x & -\sin x & 0 \\\ \sin x & \cos x & 0 \\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \\\ -\sin x & \cos x & 0 \\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{bmatrix} = I

- Thus, f(−x)f(-x) is indeed the inverse of f(x)f(x), so Statement I is true.

Step 2. Verification of Statement II: To verify f(x)⋅f(y)=f(x+y)f(x) \cdot f(y) = f(x + y), perform the matrix multiplication f(x)⋅f(y)f(x) \cdot f(y):

f(x)⋅f(y)=[cos⁡x−sin⁡x0 sin⁡xcos⁡x0 001][cos⁡y−sin⁡y0 sin⁡ycos⁡y0 001]=[cos⁡(x+y)−sin⁡(x+y)0 sin⁡(x+y)cos⁡(x+y)0 001]=f(x+y)f(x) \cdot f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\\ \sin x & \cos x & 0 \\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\\ \sin y & \cos y & 0 \\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos(x + y) & -\sin(x + y) & 0 \\\ \sin(x + y) & \cos(x + y) & 0 \\\ 0 & 0 & 1 \end{bmatrix} = f(x + y)

- Therefore, f(x)⋅f(y)=f(x+y)f(x) \cdot f(y) = f(x + y), so Statement II is also true.

Since both Statement I and Statement II are true, the correct answer is (4)(4).