atrix L nas the lemen R2 – R1) → R2, R2 + R1 pplied to it to transform it to the matrix
atrix L nas the lemen R2 – R1) → R2, R2 + R1 pplied to it to transform it to the matrix
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The matrix L has the elementary row operations ... (see image)
![### Elementary Row Operations and Determinants
The matrix \( L \) has had the elementary row operations
\[ (R_2 - R_1) \rightarrow R_2 , \quad R_2 \leftrightarrow R_1 , \quad (R_3 - 3R_2) \rightarrow R_3 \]
applied to it to transform it to the matrix:
\[ \begin{bmatrix}
1 & -1 & 1 \\
0 & 1 & -1 \\
0 & 0 & 6
\end{bmatrix} \]
**Objective:**
Calculate the determinant of the matrix \( L \).
---
**Steps to consider:**
1. **Recap Elementary Row Operations:**
- **Row Addition/Subtraction:** This operation does not change the determinant.
- **Row Swap:** This operation multiplies the determinant by \(-1\).
- **Row Multiplication:** This operation scales the determinant by the scalar.
2. **Operation Analysis:**
- **First Operation:** \((R_2 - R_1) \rightarrow R_2\)
- This is a row subtraction, which keeps the determinant unchanged.
- **Second Operation:** \(R_2 \leftrightarrow R_1\)
- This swaps rows, multiplying the determinant by \(-1\).
- **Third Operation:** \((R_3 - 3R_2) \rightarrow R_3\)
- This is a row subtraction again, which keeps the determinant unchanged.
3. **Determinant of the Transformed Matrix:**
- The transformed matrix is upper triangular:
\[ \begin{vmatrix}
1 & -1 & 1 \\
0 & 1 & -1 \\
0 & 0 & 6
\end{vmatrix} \]
- The determinant of an upper triangular matrix is the product of its diagonal elements.
- Therefore, the determinant of the initial matrix is:
\[
\text{Det} = 1 \cdot 1 \cdot 6 = 6
\]
- However, considering the row swap that occurred, we must account for the multiplication by \(-1\):
\[
\text{Final Det} = -6
\]
**Conclusion:**
The determinant](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc17e2a4-76f7-46f2-97e3-7d367a16a370%2Fe13b28d1-84ea-46fd-b9be-bf47c95a34f8%2Fv2gip9t.png&w=3840&q=75)
Transcribed Image Text:### Elementary Row Operations and Determinants
The matrix \( L \) has had the elementary row operations
\[ (R_2 - R_1) \rightarrow R_2 , \quad R_2 \leftrightarrow R_1 , \quad (R_3 - 3R_2) \rightarrow R_3 \]
applied to it to transform it to the matrix:
\[ \begin{bmatrix}
1 & -1 & 1 \\
0 & 1 & -1 \\
0 & 0 & 6
\end{bmatrix} \]
**Objective:**
Calculate the determinant of the matrix \( L \).
---
**Steps to consider:**
1. **Recap Elementary Row Operations:**
- **Row Addition/Subtraction:** This operation does not change the determinant.
- **Row Swap:** This operation multiplies the determinant by \(-1\).
- **Row Multiplication:** This operation scales the determinant by the scalar.
2. **Operation Analysis:**
- **First Operation:** \((R_2 - R_1) \rightarrow R_2\)
- This is a row subtraction, which keeps the determinant unchanged.
- **Second Operation:** \(R_2 \leftrightarrow R_1\)
- This swaps rows, multiplying the determinant by \(-1\).
- **Third Operation:** \((R_3 - 3R_2) \rightarrow R_3\)
- This is a row subtraction again, which keeps the determinant unchanged.
3. **Determinant of the Transformed Matrix:**
- The transformed matrix is upper triangular:
\[ \begin{vmatrix}
1 & -1 & 1 \\
0 & 1 & -1 \\
0 & 0 & 6
\end{vmatrix} \]
- The determinant of an upper triangular matrix is the product of its diagonal elements.
- Therefore, the determinant of the initial matrix is:
\[
\text{Det} = 1 \cdot 1 \cdot 6 = 6
\]
- However, considering the row swap that occurred, we must account for the multiplication by \(-1\):
\[
\text{Final Det} = -6
\]
**Conclusion:**
The determinant
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