2:58T00 23 == | CH 17, 1EP a. Step 1/8 The main objective is to show that 0 -{(5) с or not. Step 2/8 | b,c,de R (R, +, .) Recall that multiplication are closed. is a subring of { * b 0 A = 1- (0₂5). B = ( ) = S C d " Let can be proved as follows. .25% ال... (0) is a Ring so, addition and < M₂ (R) then the result
2:58T00 23 == | CH 17, 1EP a. Step 1/8 The main objective is to show that 0 -{(5) с or not. Step 2/8 | b,c,de R (R, +, .) Recall that multiplication are closed. is a subring of { * b 0 A = 1- (0₂5). B = ( ) = S C d " Let can be proved as follows. .25% ال... (0) is a Ring so, addition and < M₂ (R) then the result
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Why is it A-B. Isn't it A+B, since it is closes under addition. Please explain
![### CH 17, 1EP
**Step 1/8**
**Objective:**
The main goal is to determine if the set
\[
S = \left\{ \begin{pmatrix} 0 & b \\ c & d \end{pmatrix} \mid b, c, d \in \mathbb{R} \right\}
\]
is a subring of \( M_2(\mathbb{R}) \).
**Step 2/8**
**Background:**
Recall that \( (\mathbb{R}, +, \cdot) \) is a ring, meaning addition and multiplication are closed within it.
Consider matrices:
\[
A = \begin{pmatrix} 0 & b \\ c & d \end{pmatrix}, \quad B = \begin{pmatrix} 0 & e \\ f & g \end{pmatrix} \in S
\]
To test for subring properties, perform the following operations:
1. **Matrix Subtraction:**
\[
A - B = \begin{pmatrix} 0 & b - e \\ c - f & d - g \end{pmatrix}
\]
2. **Matrix Multiplication:**
\[
AB = \begin{pmatrix} b \cdot f & b \cdot g \\ d \cdot f & c \cdot e + d \cdot g \end{pmatrix}
\]
For \( AB \) to remain in \( S \), the element at position (1,1) must be 0. However, if \( b, f \neq 0 \), this condition is not satisfied (\( b \cdot f \neq 0 \)), thus \( AB \notin S \).
**Conclusion:**
Therefore,
\[
S = \left\{ \begin{pmatrix} 0 & b \\ c & d \end{pmatrix} \mid b, c, d \in \mathbb{R} \right\}
\]
is not a subring of \( M_2(\mathbb{R}) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30761ad5-6d22-4ff4-adb6-4f166a7ab52a%2Fb82e1ff5-755d-4dad-8fd7-4dd64d0eba91%2F4we23e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### CH 17, 1EP
**Step 1/8**
**Objective:**
The main goal is to determine if the set
\[
S = \left\{ \begin{pmatrix} 0 & b \\ c & d \end{pmatrix} \mid b, c, d \in \mathbb{R} \right\}
\]
is a subring of \( M_2(\mathbb{R}) \).
**Step 2/8**
**Background:**
Recall that \( (\mathbb{R}, +, \cdot) \) is a ring, meaning addition and multiplication are closed within it.
Consider matrices:
\[
A = \begin{pmatrix} 0 & b \\ c & d \end{pmatrix}, \quad B = \begin{pmatrix} 0 & e \\ f & g \end{pmatrix} \in S
\]
To test for subring properties, perform the following operations:
1. **Matrix Subtraction:**
\[
A - B = \begin{pmatrix} 0 & b - e \\ c - f & d - g \end{pmatrix}
\]
2. **Matrix Multiplication:**
\[
AB = \begin{pmatrix} b \cdot f & b \cdot g \\ d \cdot f & c \cdot e + d \cdot g \end{pmatrix}
\]
For \( AB \) to remain in \( S \), the element at position (1,1) must be 0. However, if \( b, f \neq 0 \), this condition is not satisfied (\( b \cdot f \neq 0 \)), thus \( AB \notin S \).
**Conclusion:**
Therefore,
\[
S = \left\{ \begin{pmatrix} 0 & b \\ c & d \end{pmatrix} \mid b, c, d \in \mathbb{R} \right\}
\]
is not a subring of \( M_2(\mathbb{R}) \).
Expert Solution
Step 1: Given a subset S of 2×2 matrices over reals
To check: S is a subring or not
Step by step
Solved in 4 steps with 2 images
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