VIPSolutionsGive step-by-step solution with explanation and final answer:2(a) For the arbitrary vector (x y 2), the operators E, Cz, 6, and 6, can be described by the following 3x3 matrices: 100 -1 0 0 1 0 0 -1 0 0 Eo 10 Czlo -10 6:10 1 0 (0 10 0 0 1 0 0 1 0 0 1 0 01 Evaluate C26, and C2C2. What operation is equivalent to each of the two sequential operations? (6) 2 (b) Combine the following matrices: 2 0 0]px 100 H = 3) 0 0 4llz
Question:
Give step-by-step solution with explanation and final answer:
2(a) For the arbitrary vector (x y 2), the operators E, Cz, 6, and 6, can be described by the following
3x3 matrices:
100 -1 0 0 1 0 0 -1 0 0
Eo 10 Czlo -10 6:10 1 0 (0 10
0 0 1 0 0 1 0 0 1 0 01
Evaluate C26, and C2C2. What operation is equivalent to each of the two sequential operations?
(6)
2 (b) Combine the following matrices:
2 0 0]px
100 H = 3)
0 0 4llz
This is AI generated Response.
Asked by: 66
Created at: 2025-06-02 14:24:46
Asked by: 66
Created at: 2025-06-02 14:24:46
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Let's solve each part step-by-step:
---
## **2(a)**
### **Given Matrices:**
\[
\mathbf{E} =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\quad
\mathbf{C}_2 =
\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
\[
\hat{\sigma}_v =
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\quad
\hat{\sigma}_v' =
\begin{bmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
---
### **Evaluate \( \hat{C}_2 \hat{\sigma}_v \) and \( \hat{C}_2 \hat{C}_2 \):**
#### **1. \( \hat{C}_2 \hat{\sigma}_v \):**
Multiply \( \hat{C}_2 \) by \( \hat{\sigma}_v \):
\[
\hat{C}_2 \hat{\sigma}_v =
\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
Perform the matrix multiplication:
- Row 1: \((-1*1) + (0*0) + (0*0) = -1\), \((-1*0) + (0*-1) + (0*0) = 0\), \((-1*0) + (0*0) + (0*1) = 0\)
- Row 2: \((0*1) + (-1*0) + (0*0) = 0\), \((0*0) + (-1*-1) + (0*0) = 1\), \((0*0) + (-1*0) + (0*1) = 0\)
- Row 3: \((0*1) + (0*0) + (1*0) = 0\), \((0*0) + (0*-1) + (1*0) = 0\), \((0*0) + (0*0) + (1*1) = 1\)
So,
\[
\hat{C}_2 \hat{\sigma}_v =
\begin{bmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
= \hat{\sigma}_v'
\]
**Operation Equivalent:** This is equivalent to applying the reflection \( \hat{\sigma}_v' \).
---
#### **2. \( \hat{C}_2 \hat{C}_2 \):**
Multiply \( \hat{C}_2 \) by itself:
\[
\hat{C}_2 \hat{C}_2 =
\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
- Row 1: \((-1*-1) + (0*0) + (0*0) = 1\), \((-1*0) + (0*-1) + (0*0) = 0\), \((-1*0) + (0*0) + (0*1) = 0\)
- Row 2: \((0*-1) + (-1*0) + (0*0) = 0\), \((0*0) + (-1*-1) + (0*0) = 1\), \((0*0) + (-1*0) + (0*1) = 0\)
- Row 3: \((0*-1) + (0*0) + (1*0) = 0\), \((0*0) + (0*-1) + (1*0) = 0\), \((0*0) + (0*0) + (1*1) = 1\)
So,
\[
\hat{C}_2 \hat{C}_2 =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
= \mathbf{E}
\]
**Operation Equivalent:** This is equivalent to the identity operation (no change).
---
## **2(b)**
Multiply the matrix with the vector:
\[
\begin{bmatrix}
2 & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 4
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
2x \\
x \\
4z
\end{bmatrix}
\]
---
## **Summary of Answers**
### **2(a):**
- \( \hat{C}_2 \hat{\sigma}_v = \hat{\sigma}_v' \) (reflection through the y-z plane)
- \( \hat{C}_2 \hat{C}_2 = \mathbf{E} \) (identity operation)
### **2(b):**
\[
\begin{bmatrix}
2x \\
x \\
4z
\end{bmatrix}
\]
---
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