Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter3: Matrices
Section3.1: Matrix Operations
Problem 20EQ: Referring to Exercise 19, suppose that the unit cost of distributing the products to stores is the...
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Question
12,13
![**Perform the Row Operation on the Given Matrix**
**Objective:**
Perform the row operation \( \frac{1}{2}R_1 \rightarrow R_1 \) on the matrix below.
**Given Matrix:**
\[
\begin{bmatrix}
-4 & 6 & \mid & -8 \\
-4 & 5 & \mid & 3 \\
\end{bmatrix}
\]
**Operation:**
\[ \frac{1}{2}R_1 \rightarrow R_1 \]
This means the first row \( R_1 \) of the matrix will be multiplied by \( \frac{1}{2} \).
**Step-by-Step Solution:**
1. **Multiply Each Element in \( R_1 \) by \( \frac{1}{2} \):**
- First element: \( -4 \times \frac{1}{2} = -2 \)
- Second element: \( 6 \times \frac{1}{2} = 3 \)
- Third element (after the vertical line): \( -8 \times \frac{1}{2} = -4 \)
2. **Form the New Matrix:**
\[
\begin{bmatrix}
-2 & 3 & \mid & -4 \\
-4 & 5 & \mid & 3 \\
\end{bmatrix}
\]
**Simplified Answer:**
\[
\begin{bmatrix}
-2 & 3 & \mid & -4 \\
-4 & 5 & \mid & 3 \\
\end{bmatrix}
\]
This completes the row operation on the given matrix.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a1af21b-af18-455d-9657-67bfb10798dd%2Fbcd0638c-9b8f-4742-9338-beb806061dc0%2Feyjfuu_processed.png&w=3840&q=75)
Transcribed Image Text:**Perform the Row Operation on the Given Matrix**
**Objective:**
Perform the row operation \( \frac{1}{2}R_1 \rightarrow R_1 \) on the matrix below.
**Given Matrix:**
\[
\begin{bmatrix}
-4 & 6 & \mid & -8 \\
-4 & 5 & \mid & 3 \\
\end{bmatrix}
\]
**Operation:**
\[ \frac{1}{2}R_1 \rightarrow R_1 \]
This means the first row \( R_1 \) of the matrix will be multiplied by \( \frac{1}{2} \).
**Step-by-Step Solution:**
1. **Multiply Each Element in \( R_1 \) by \( \frac{1}{2} \):**
- First element: \( -4 \times \frac{1}{2} = -2 \)
- Second element: \( 6 \times \frac{1}{2} = 3 \)
- Third element (after the vertical line): \( -8 \times \frac{1}{2} = -4 \)
2. **Form the New Matrix:**
\[
\begin{bmatrix}
-2 & 3 & \mid & -4 \\
-4 & 5 & \mid & 3 \\
\end{bmatrix}
\]
**Simplified Answer:**
\[
\begin{bmatrix}
-2 & 3 & \mid & -4 \\
-4 & 5 & \mid & 3 \\
\end{bmatrix}
\]
This completes the row operation on the given matrix.
![### Matrix Row Operation Example
**Problem Statement:**
Perform the row operation \( R_1 + R_2 \rightarrow R_1 \) on the matrix below.
\[
\begin{bmatrix}
4 & -6 & 7 \\
3 & -5 & 2
\end{bmatrix}
\]
**Application of the Row Operation:**
\[
\begin{bmatrix}
4 & -6 & 7 \\
3 & -5 & 2
\end{bmatrix}
\begin{array}{c}
R_1 + R_2 \rightarrow R_1
\end{array}
\sim
\begin{bmatrix}
[ \boxed{} & \boxed{} & \boxed{} ] \\
[ 3 & -5 & 2 ]
\end{bmatrix}
\]
**Instructions:**
Simplify your answer by calculating the new entries in the first row after applying the row operation \( R_1 + R_2 \rightarrow R_1 \). Replace the boxed placeholders with the calculated values.
**Detailed Steps:**
1. Add the corresponding elements of \( R_1 \) and \( R_2 \):
- First element: \( 4 + 3 = 7 \)
- Second element: \( -6 + (-5) = -11 \)
- Third element: \( 7 + 2 = 9 \)
2. Update the first row with the new values:
\[
\begin{bmatrix}
[ 7 & -11 & 9 ] \\
[ 3 & -5 & 2 ]
\end{bmatrix}
\]
The final matrix after performing the row operation \( R_1 + R_2 \rightarrow R_1 \) is:
\[
\begin{bmatrix}
7 & -11 & 9 \\
3 & -5 & 2
\end{bmatrix}
\]
**Summary:**
Applying the row operation helps in simplifying matrices for further mathematical operations, especially in solving systems of linear equations using methods like Gaussian elimination.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a1af21b-af18-455d-9657-67bfb10798dd%2Fbcd0638c-9b8f-4742-9338-beb806061dc0%2F8uil4ei_processed.png&w=3840&q=75)
Transcribed Image Text:### Matrix Row Operation Example
**Problem Statement:**
Perform the row operation \( R_1 + R_2 \rightarrow R_1 \) on the matrix below.
\[
\begin{bmatrix}
4 & -6 & 7 \\
3 & -5 & 2
\end{bmatrix}
\]
**Application of the Row Operation:**
\[
\begin{bmatrix}
4 & -6 & 7 \\
3 & -5 & 2
\end{bmatrix}
\begin{array}{c}
R_1 + R_2 \rightarrow R_1
\end{array}
\sim
\begin{bmatrix}
[ \boxed{} & \boxed{} & \boxed{} ] \\
[ 3 & -5 & 2 ]
\end{bmatrix}
\]
**Instructions:**
Simplify your answer by calculating the new entries in the first row after applying the row operation \( R_1 + R_2 \rightarrow R_1 \). Replace the boxed placeholders with the calculated values.
**Detailed Steps:**
1. Add the corresponding elements of \( R_1 \) and \( R_2 \):
- First element: \( 4 + 3 = 7 \)
- Second element: \( -6 + (-5) = -11 \)
- Third element: \( 7 + 2 = 9 \)
2. Update the first row with the new values:
\[
\begin{bmatrix}
[ 7 & -11 & 9 ] \\
[ 3 & -5 & 2 ]
\end{bmatrix}
\]
The final matrix after performing the row operation \( R_1 + R_2 \rightarrow R_1 \) is:
\[
\begin{bmatrix}
7 & -11 & 9 \\
3 & -5 & 2
\end{bmatrix}
\]
**Summary:**
Applying the row operation helps in simplifying matrices for further mathematical operations, especially in solving systems of linear equations using methods like Gaussian elimination.
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