1. Key Concepts and Formulas

• Matrix Multiplication: To multiply two matrices $A$ (of dimension $m \times n$) and $B$ (of dimension $n \times p$), the element in the $i$-th row and $j$-th column of the product $AB$ is obtained by taking the dot product of the $i$-th row of $A$ and the $j$-th column of $B$.
• Identity Matrix ($I$): A square matrix with ones on the main diagonal and zeros elsewhere. For a $3 \times 3$ matrix, $I = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$.
• Inverse of a Matrix: For a square matrix $A$, its inverse $A^{-1}$ (if it exists) is a matrix such that $AA^{-1} = A^{-1}A = I$. A matrix $A$ is invertible if and only if its determinant, $\det(A)$, is non-zero.
• Trigonometric Identities:
  • $\cos(A+B) = \cos A \cos B - \sin A \sin B$
  • $\sin(A+B) = \sin A \cos B + \cos A \sin B$
  • $\cos(-x) = \cos x$
  • $\sin(-x) = -\sin x$
  • $\cos^2 x + \sin^2 x = 1$

2. Step-by-Step Solution

Let the given matrix be $f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$.

Step 1: Evaluate Statement II: $f(x)f(y) = f(x+y)$.

• What we are doing: We need to compute the matrix product $f(x)f(y)$ and compare it with $f(x+y)$.
• Why: This will determine the truth value of Statement II. This property is also useful for evaluating Statement I.
• Calculation: We have $f(x) = \left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$ and $f(y) = \left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$. Now, let's calculate their product $f(x)f(y)$: $f(x)f(y) = \left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right] \left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$ Performing the matrix multiplication:
  • $(1,1)$ element: $(\cos x)(\cos y) + (-\sin x)(\sin y) + (0)(0) = \cos x \cos y - \sin x \sin y = \cos(x+y)$
  • $(1,2)$ element: $(\cos x)(-\sin y) + (-\sin x)(\cos y) + (0)(0) = -(\cos x \sin y + \sin x \cos y) = -\sin(x+y)$
  • $(1,3)$ element: $(\cos x)(0) + (-\sin x)(0) + (0)(1) = 0$
  • $(2,1)$ element: $(\sin x)(\cos y) + (\cos x)(\sin y) + (0)(0) = \sin x \cos y + \cos x \sin y = \sin(x+y)$
  • $(2,2)$ element: $(\sin x)(-\sin y) + (\cos x)(\cos y) + (0)(0) = \cos x \cos y - \sin x \sin y = \cos(x+y)$
  • $(2,3)$ element: $(\sin x)(0) + (\cos x)(0) + (0)(1) = 0$
  • $(3,1)$ element: $(0)(\cos y) + (0)(\sin y) + (1)(0) = 0$
  • $(3,2)$ element: $(0)(-\sin y) + (0)(\cos y) + (1)(0) = 0$
  • $(3,3)$ element: $(0)(0) + (0)(0) + (1)(1) = 1$ Thus, we get: $f(x)f(y) = \left[\begin{array}{ccc}\cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1\end{array}\right]$ By definition, this is exactly $f(x+y)$.
• Reasoning: The result matches the definition of $f(x+y)$.
• Conclusion for Statement II: Statement II is True.

Step 2: Evaluate Statement I: $f(-x)$ is the inverse of the matrix $f(x)$.

• What we are doing: To check if $f(-x)$ is the inverse of $f(x)$, we need to verify if their product is the identity matrix, i.e., $f(x)f(-x) = I$. We can use the property established in Statement II.
• Why: If the product is the identity matrix, then $f(-x)$ is the inverse of $f(x)$.
• Calculation: From Statement II, we know that $f(A)f(B) = f(A+B)$. Let's substitute $A=x$ and $B=-x$ into this property: $f(x)f(-x) = f(x+(-x)) = f(0)$ Now, let's find the matrix $f(0)$ by substituting $x=0$ into the definition of $f(x)$: $f(0) = \left[\begin{array}{ccc}\cos 0 & -\sin 0 & 0 \\ \sin 0 & \cos 0 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ This is the $3 \times 3$ identity matrix, $I$. Therefore, $f(x)f(-x) = I$. To fully confirm it's the inverse, we should also check $f(-x)f(x)=I$. Using the same property: $f(-x)f(x) = f(-x+x) = f(0) = I$. Since both $f(x)f(-x) = I$ and $f(-x)f(x) = I$, $f(-x)$ is indeed the inverse of $f(x)$.
• Reasoning: The product of $f(x)$ and $f(-x)$ (in both orders) results in the identity matrix, which is the definition of an inverse.
• Conclusion for Statement I: Statement I is True.

Step 3: Combine conclusions.

• Statement I is True.
• Statement II is True.

Therefore, both Statement I and Statement II are true.

3. Common Mistakes & Tips

• Matrix Multiplication Errors: Be meticulous with the dot products for each element. A single sign error or incorrect element can lead to a wrong conclusion.
• Trigonometric Identities: Incorrectly applying $\cos(-x) = \cos x$ or $\sin(-x) = -\sin x$ can lead to errors when evaluating $f(-x)$.
• Definition of Inverse: Remember that for $B$ to be the inverse of $A$, both $AB=I$ and $BA=I$ must hold. For invertible square matrices, if one holds, the other usually follows, but it's good practice to be aware.
• Recognizing Standard Forms: The matrix $f(x)$ is a standard rotation matrix about the z-axis. Its properties, $f(x)f(y) = f(x+y)$ and $f(x)^{-1} = f(-x)$, are fundamental for rotation matrices.

4. Summary

We systematically evaluated both statements. For Statement II, we performed matrix multiplication of $f(x)$ and $f(y)$ and used trigonometric sum identities to show that the result is indeed $f(x+y)$, confirming Statement II as true. For Statement I, we leveraged the property established in Statement II to calculate $f(x)f(-x)$, which simplified to $f(0)$. Evaluating $f(0)$ yielded the identity matrix, proving that $f(-x)$ is the inverse of $f(x)$. Thus, Statement I is also true. Both statements are true.

5. Final Answer

Based on our analysis, both Statement I and Statement II are true. The final answer is \boxed{C}