VIPSolutionsgive answer in 3 step with explanation at the end of each step and final answer at the end:Question 2 [L] 2(a) For the arbitrary vector (x y 2), the operators E, C2, 8, and 8,’ can be described by the following 3%3 matrices: 100 -1 0 0 1.0 0 ~1 0 0 ef 1 9 eo -1 9 alo -1 ol ao 1 o 0 01 0 0 1 0 0 1 001 Exélasts Caf, and Cie: What opatiiion & e.évslect kx ach of 16 wo sequenbal cpurations? 6)
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give answer in 3 step with explanation at the end of each step and final answer at the end:
Question 2 [L]
2(a) For the arbitrary vector (x y 2), the operators E, C2, 8, and 8,’ can be described by the following
3%3 matrices:
100 -1 0 0 1.0 0 ~1 0 0
ef 1 9 eo -1 9 alo -1 ol ao 1 o
0 01 0 0 1 0 0 1 001
Exélasts Caf, and Cie: What opatiiion & e.évslect kx ach of 16 wo sequenbal cpurations?
6)
This is AI generated Response.
Asked by: saravjeet
Created at: 2025-05-25 19:48:00
Asked by: saravjeet
Created at: 2025-05-25 19:48:00
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Let's break down the problem in three steps for **each sequential operation**:
---
### **Step 1: Write Down the Matrices**
**Given Matrices:**
\[
\hat{E} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\]
\[
\hat{C}_2 = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\]
\[
\hat{\sigma}_v = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\]
\[
\hat{\sigma}_v' = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\]
**Explanation:**
These are the symmetry operation matrices for a 3D vector \((x, y, z)\).
---
### **Step 2: Evaluate \(\hat{C}_2 \hat{\sigma}_v'\) and \(\hat{C}_2 \hat{C}_2\)**
#### **2.1: Calculate \(\hat{C}_2 \hat{\sigma}_v'\)**
\[
\hat{C}_2 = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\]
\[
\hat{\sigma}_v' = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\]
Multiply them:
\[
\hat{C}_2 \hat{\sigma}_v' =
\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\]
- Row 1: (-1)\*(-1) = 1, rest 0
- Row 2: (-1)\*0 + 1\*1 = -1
- Row 3: 1\*1 = 1
So:
\[
\hat{C}_2 \hat{\sigma}_v' =
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
**Explanation:**
Multiplying the two matrices gives us a new transformation matrix.
---
#### **2.2: Calculate \(\hat{C}_2 \hat{C}_2\)**
\[
\hat{C}_2 \hat{C}_2 =
\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
=
\begin{bmatrix}
(-1)\*(-1) & 0 & 0 \\
0 & (-1)\*(-1) & 0 \\
0 & 0 & 1*1
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
**Explanation:**
Multiplying \(\hat{C}_2\) by itself gives the identity matrix, meaning doing the operation twice brings the system back to the start.
---
### **Step 3: Identify the Equivalent Operations**
- \(\hat{C}_2 \hat{\sigma}_v' = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \hat{\sigma}_v\)
- \(\hat{C}_2 \hat{C}_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \hat{E}\)
**Explanation:**
- \(\hat{C}_2 \hat{\sigma}_v'\) is equivalent to the \(\sigma_v\) reflection.
- \(\hat{C}_2 \hat{C}_2\) is equivalent to the identity operation \(\hat{E}\).
---
## **Final Answers**
1. **\(\hat{C}_2 \hat{\sigma}_v'\) is equivalent to the mirror operation \(\hat{\sigma}_v\).**
2. **\(\hat{C}_2 \hat{C}_2\) is equivalent to the identity operation \(\hat{E}\).**
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