**Problem Statement (Linear Algebra):****Prove WITHOUT using the product property of determinants: If $ M $, $ N $ are triangular matrices, $ |MN| = |M||N| $.**(Note: The provided image includes text that is partially obscured and thus incomplete. The essential statement remains to prove that the determinant of the product of two triangular matrices is equal to the product of their determinants, but without relying on the known property of determinants.)---### Detailed ExplanationIn linear algebra, a triangular matrix is a special kind of square matrix. There are two types of triangular matrices:1. **Upper triangular matrix**: All the entries below the main diagonal are zero.2. **Lower triangular matrix**: All the entries above the main diagonal are zero.### Key Points to Cover in the Proof:1. **Basics of Triangular Matrices**: Explain the definitions, properties, and characteristics of upper and lower triangular matrices.2. **Determinants of Triangular Matrices**: Describe how the determinant of a triangular matrix is the product of its diagonal elements.3. **Product of Two Triangular Matrices**: - Show how the product of two upper triangular matrices is upper triangular. - Show how the product of two lower triangular matrices is lower triangular.4. **Diagonal Elements**: Focus on the diagonal elements of $ MN $ and how they relate to the diagonal elements of $ M $ and $ N $.5. **Conclusion**: Prove that $ |MN| = |M||N| $ by considering the product of the diagonal elements of the resulting triangular matrix from the multiplication $ MN $. ### Potential Steps in the Proof:- Analyze elements at positions (i, i) in the matrix $ MN $.- Since (MN)_{ii} = $ M_{ii}N_{ii} $, the determinant formula follows the product of these diagonal elements for both matrices $ M $ and $ N $.By carefully walking through these steps, the proof will show the result without directly invoking the property $ |MN| = |M||N| $ outright, thus adhering to the problem's constraints.

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Answered: Prove WITHOUT using the product…

Math

Advanced Math

Prove WITHOUT using the product property of determinants: If M, N are triangular matrices, |MN| = |M||N].

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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Question

**Problem Statement (Linear Algebra):**

**Prove WITHOUT using the product property of determinants: If $ M $, $ N $ are triangular matrices, $ |MN| = |M||N| $.**

(Note: The provided image includes text that is partially obscured and thus incomplete. The essential statement remains to prove that the determinant of the product of two triangular matrices is equal to the product of their determinants, but without relying on the known property of determinants.)

---

### Detailed Explanation

In linear algebra, a triangular matrix is a special kind of square matrix. There are two types of triangular matrices:
1. **Upper triangular matrix**: All the entries below the main diagonal are zero.
2. **Lower triangular matrix**: All the entries above the main diagonal are zero.

### Key Points to Cover in the Proof:
1. **Basics of Triangular Matrices**: Explain the definitions, properties, and characteristics of upper and lower triangular matrices.
2. **Determinants of Triangular Matrices**: Describe how the determinant of a triangular matrix is the product of its diagonal elements.
3. **Product of Two Triangular Matrices**:
- Show how the product of two upper triangular matrices is upper triangular.
- Show how the product of two lower triangular matrices is lower triangular.
4. **Diagonal Elements**: Focus on the diagonal elements of $ MN $ and how they relate to the diagonal elements of $ M $ and $ N $.
5. **Conclusion**: Prove that $ |MN| = |M||N| $ by considering the product of the diagonal elements of the resulting triangular matrix from the multiplication $ MN $.

### Potential Steps in the Proof:
- Analyze elements at positions (i, i) in the matrix $ MN $.
- Since (MN)_{ii} = $ M_{ii}N_{ii} $, the determinant formula follows the product of these diagonal elements for both matrices $ M $ and $ N $.

By carefully walking through these steps, the proof will show the result without directly invoking the property $ |MN| = |M||N| $ outright, thus adhering to the problem's constraints.

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