you will receive a question-answer pair and be asked to rewrite just  the answer and explains your reasoning step by step, incorporating dynamic Chain of Thought (CoT), reflection, and verbal reinforcement learning. Follow these instructions:

1. Enclose all thoughts within <thinking> tags, exploring multiple angles and approaches.
2. Break down the solution into clear steps, providing a title and content for each step.
3. After each step, decide if you need another step or if you're ready to give the final answer.
4. Continuously adjust your reasoning based on intermediate results and reflections, adapting your strategy as you progress.
5. Regularly evaluate your progress, being critical and honest about your reasoning process.
6. Assign a confidence score between 0.0 and 1.0 to guide your approach:
   - 0.8+: Continue current approach
   - 0.5-0.7: Consider minor adjustments
   - Below 0.5: Seriously consider backtracking and trying a different approach
7. If unsure or if your score is low, backtrack and try a different approach, explaining your decision.
8. For mathematical problems, show all work explicitly using LaTeX for formal notation and provide detailed proofs.
9. Explore multiple solutions individually if possible, comparing approaches in your reflections.
10. Use your thoughts as a scratchpad, writing out all calculations and reasoning explicitly.
11. Use at least 5 methods to derive the answer and consider alternative viewpoints.
12. Be aware of your limitations as an AI and what you can and cannot do.

### Example Problem:
*Problem*: Solve for \( x \) in the equation \( 2x + 3 = 11 \).

### Expected Format:
*Step 1*: Subtract 3 from both sides of the equation.  
<thinking>Subtracting 3 from both sides should isolate the term with \( x \). Am I confident this is the correct first step? Yes, because it simplifies the equation to \( 2x = 8 \).</thinking>  
*Step 1 Confidence Score*: 0.85

*Step 2*: Divide both sides by 2.  
<thinking>Dividing by 2 will solve for \( x \). This is a straightforward division. Is there any reason to doubt this step? No, it should give \( x = 4 \).</thinking>  
*Step 2 Confidence Score*: 0.90

*Step 3*: Verify the solution by substituting \( x = 4 \) back into the original equation.  
<thinking>Substituting \( x = 4 \) back into the equation should confirm the solution. Does \( 2(4) + 3 = 11 \)? Yes, it does, so the solution is verified.</thinking>  
*Step 3 Confidence Score*: 1.00

Question: Find the matrix $\mathbf{R}$ such that for any vector $\mathbf{v}$, $\mathbf{R} \mathbf{v}$ is the reflection of $\mathbf{v}$ across the line $L$ defined by the equation $\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

This problem requires finding the reflection matrix across a line defined by a vector equation, rather than a plane defined by a normal vector. The approach to solve this problem would involve using the cross product to find the projection of $\mathbf{v}$ onto the line, and then using the projection to construct the reflection matrix.
Answer: First, find the unit vector $\mathbf{u}$ along the line $L$. Since $L$ is defined by the equation $\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$, the direction vector of $L$ is $\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$. Normalize this vector to get $\mathbf{u}$:
\[ \mathbf{u} = \frac{\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}}{\sqrt{6}} \]

The projection matrix $\mathbf{P}$ that projects any vector onto $L$ is given by $\mathbf{P} = \mathbf{u} \mathbf{u}^T$:
\[ \mathbf{P} = \frac{1}{6} \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \begin{pmatrix} 1 & 2 & -1 \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \\ -1 & -2 & 1 \end{pmatrix} \]

The reflection matrix $\mathbf{R}$ across the line $L$ can be found using the formula $\mathbf{R} = 2\mathbf{P} - \mathbf{I}$, where $\mathbf{I}$ is the identity matrix:
\[ \mathbf{R} = 2\mathbf{P} - \mathbf{I} = \frac{1}{3} \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \\ -1 & -2 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} -2 & 2 & -1 \\ 2 & 1 & -2 \\ -1 & -2 & -2 \end{pmatrix} \]

Therefore, the reflection matrix $\mathbf{R}$ is:
\[ \mathbf{R} = \boxed{\begin{pmatrix} -\frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ -\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3} \end{pmatrix}} \]