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Question: If $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$ t...

If f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} then f(x+y)f(x+y) is equal to:

A

f(x)+f(y)

B

f(x)-f(y)

C

f(x) \cdot f(y)

D

None of these

Answer

f(x) \cdot f(y)

Explanation

Solution

To find f(x+y)f(x+y), we replace xx with x+yx+y in the definition of f(x)f(x): f(x+y)=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0001]f(x+y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}.

Now, let's compute the product f(x)⋅f(y)f(x) \cdot f(y): f(x)⋅f(y)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001][cos⁡y−sin⁡y0sin⁡ycos⁡y0001]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}

Performing matrix multiplication: The element in the first row, first column is (cos⁡x)(cos⁡y)+(−sin⁡x)(sin⁡y)+(0)(0)=cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y)(\cos x)(\cos y) + (-\sin x)(\sin y) + (0)(0) = \cos x \cos y - \sin x \sin y = \cos(x+y). The element in the first row, second column is (cos⁡x)(−sin⁡y)+(−sin⁡x)(cos⁡y)+(0)(0)=−cos⁡xsin⁡y−sin⁡xcos⁡y=−(sin⁡xcos⁡y+cos⁡xsin⁡y)=−sin⁡(x+y)(\cos x)(-\sin y) + (-\sin x)(\cos y) + (0)(0) = -\cos x \sin y - \sin x \cos y = -(\sin x \cos y + \cos x \sin y) = -\sin(x+y). The element in the second row, first column is (sin⁡x)(cos⁡y)+(cos⁡x)(sin⁡y)+(0)(0)=sin⁡xcos⁡y+cos⁡xsin⁡y=sin⁡(x+y)(\sin x)(\cos y) + (\cos x)(\sin y) + (0)(0) = \sin x \cos y + \cos x \sin y = \sin(x+y). The element in the second row, second column is (sin⁡x)(−sin⁡y)+(cos⁡x)(cos⁡y)+(0)(0)=−sin⁡xsin⁡y+cos⁡xcos⁡y=cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y)(\sin x)(-\sin y) + (\cos x)(\cos y) + (0)(0) = -\sin x \sin y + \cos x \cos y = \cos x \cos y - \sin x \sin y = \cos(x+y). The third row and third column elements remain unchanged, resulting in [001]\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}.

Thus, f(x)⋅f(y)=[cos⁡(x+y)−sin⁡(x+y)0sin⁡(x+y)cos⁡(x+y)0001]f(x) \cdot f(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}. This is equal to f(x+y)f(x+y).