Assume A and B are both n × n matrices. If det A = 2, then the determinant of –BAB¯' is
Assume A and B are both n × n matrices. If det A = 2, then the determinant of –BAB¯' is
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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![**Problem Statement:**
1. Assume \( A \) and \( B \) are both \( n \times n \) matrices. If \( \det A = 2 \), then the determinant of \( -BAB^{-1} \) is ______.
**Explanation:**
- We start with the knowledge that \( A \) and \( B \) are square matrices of size \( n \times n \).
- The determinant of matrix \( A \) is given as 2.
- We need to determine the determinant of the expression \( -BAB^{-1} \).
**Solution Outline:**
1. **Matrix Multiplication and Determinants:** The determinant of a product of matrices can be expressed as the product of their determinants. Specifically, for any \( n \times n \) matrices \( A \) and \( B \):
\[ \det(AB) = \det(A) \cdot \det(B) \]
2. **Properties of Determinants:**
- The determinant of the inverse of a matrix \( B \), \( \det(B^{-1}) \), is the reciprocal of \( \det(B) \).
- For a scalar multiplication of a matrix \( cA \), where \( c \) is a scalar and \( A \) is an \( n \times n \) matrix, the determinant is given by \( \det(cA) = c^n \det(A) \).
3. **Applying these Properties to the Problem:**
- Given the expression \( -BAB^{-1} \):
\[ \det(-BAB^{-1}) \]
- Applying the properties of determinants, we use:
\[ \det(-BAB^{-1}) = \det(-1 \cdot BAB^{-1}) \]
- Separating the scalar multiplication (since \(-1\) can be considered as a scalar matrix \( -I \) with \( I \) being the identity matrix):
\[ \det(-1) \cdot \det(BAB^{-1}) \]
- \( \det(-1) \), for an \( n \times n \) matrix, is \((-1)^n \)
- Now, using that \( B \cdot A \cdot B^{-1} \):
\[ \det(BAB^{-1}) = \det(B) \cdot \det(A) \cdot \det(B^{-1}) \]
- Simplifying:
\[ = \det](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1725b614-c61a-470a-b05f-008b5304bcc7%2F0a51cc41-949c-44e1-b008-461871b65ea1%2Fak2t9eo_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
1. Assume \( A \) and \( B \) are both \( n \times n \) matrices. If \( \det A = 2 \), then the determinant of \( -BAB^{-1} \) is ______.
**Explanation:**
- We start with the knowledge that \( A \) and \( B \) are square matrices of size \( n \times n \).
- The determinant of matrix \( A \) is given as 2.
- We need to determine the determinant of the expression \( -BAB^{-1} \).
**Solution Outline:**
1. **Matrix Multiplication and Determinants:** The determinant of a product of matrices can be expressed as the product of their determinants. Specifically, for any \( n \times n \) matrices \( A \) and \( B \):
\[ \det(AB) = \det(A) \cdot \det(B) \]
2. **Properties of Determinants:**
- The determinant of the inverse of a matrix \( B \), \( \det(B^{-1}) \), is the reciprocal of \( \det(B) \).
- For a scalar multiplication of a matrix \( cA \), where \( c \) is a scalar and \( A \) is an \( n \times n \) matrix, the determinant is given by \( \det(cA) = c^n \det(A) \).
3. **Applying these Properties to the Problem:**
- Given the expression \( -BAB^{-1} \):
\[ \det(-BAB^{-1}) \]
- Applying the properties of determinants, we use:
\[ \det(-BAB^{-1}) = \det(-1 \cdot BAB^{-1}) \]
- Separating the scalar multiplication (since \(-1\) can be considered as a scalar matrix \( -I \) with \( I \) being the identity matrix):
\[ \det(-1) \cdot \det(BAB^{-1}) \]
- \( \det(-1) \), for an \( n \times n \) matrix, is \((-1)^n \)
- Now, using that \( B \cdot A \cdot B^{-1} \):
\[ \det(BAB^{-1}) = \det(B) \cdot \det(A) \cdot \det(B^{-1}) \]
- Simplifying:
\[ = \det
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