Problem 17 Show that \(\operatorname{det}\l... [FREE SOLUTION]

Chapter 3: Problem 17

Show that \(\operatorname{det}\left(A+B^{T}\right)=\operatorname{det}\left(A^{T}+B\right)\) for any \(n \times n\) matrices \(A\) and \(B\).

Short Answer

Expert verified

The determinants are equal due to the property that \( \operatorname{det}(X) = \operatorname{det}(X^T) \).

Step by step solution

Understand the Determinant Property

The determinant of a matrix has a fundamental property: for any square matrix, the determinant of the transpose of a matrix is equal to the determinant of the matrix itself. That is, \( \operatorname{det}(A^T) = \operatorname{det}(A) \). This property will be critical in proving the statement.

Apply the Transpose Property

We need to find the determinant of \( A + B^T \) and compare it to the determinant of \( A^T + B \). Notice that the determinant of a transpose is equal to the determinant of the original matrix. Hence, \( \operatorname{det}((A + B^T)^T) = \operatorname{det}(A + B^T) \).

Simplify \( (A + B^T)^T \)

The transpose of a sum of matrices is equal to the sum of their transposes. Therefore, \( (A + B^T)^T = A^T + (B^T)^T = A^T + B \). This step shows us that \( A + B^T \) and \( A^T + B \) are transposes of each other.

Use the Determinant Transpose Property

Since \( (A + B^T)^T = A^T + B \), and using the property from Step 1 that \( \operatorname{det}(X^T) = \operatorname{det}(X) \), we derive that \( \operatorname{det}(A + B^T) = \operatorname{det}((A + B^T)^T) = \operatorname{det}(A^T + B) \).

Conclusion: Prove the Statement

Since we have shown that \( \operatorname{det}(A + B^T) = \operatorname{det}(A^T + B) \), the given statement is true for any \( n \times n \) matrices \( A \) and \( B \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transpose of a Matrix

In mathematics, especially in linear algebra, the transpose of a matrix is a common concept. Transposing a matrix involves flipping it over its diagonal. This swaps the matrix's row and column indices.
To be more specific:

If you have a matrix A, its transpose is denoted as A^T.
An element in the i-th row and j-th column of A becomes the element in the j-th row and i-th column of A^T.

Understanding how to transpose a matrix is fundamental in proving matrix-related properties, like the one involving determinant equivalences. It plays a vital role in the step-by-step solution provided in the original exercise.
For example, if you have the transpose operation on a sum of matrices, (A + B^T)^T = A^T + (B^T)^T, the transposing is applied to each matrix individually then added.

Square Matrix

A square matrix is a type of matrix that is widely used in mathematics. It is characterized by having the same number of rows and columns.
This means that if the order of a square matrix is n, it has n rows and n columns.

Square matrices are often used because their properties allow for operations like calculating determinants and eigenvalues.
They are essential in defining identity matrices and in performing linear transformations.

A neat property of square matrices is that you can easily find their determinant, which is a critical concept when dealing with matrices. The determinant of a square matrix influences many of its properties and abilities.
In the context of the original exercise, square matrices A and B are fundamental because their equal size allows for operations such as addition and transposing.

Properties of Determinants

Determinants provide important insights into a square matrix. They are a special number calculated from a square matrix, and they can tell you a lot about the matrix's capabilities.

One of the key properties is that the determinant of a transpose of a matrix is equal to the determinant of the original matrix. So, det(A^T) = det(A).
The determinant is useful for determining if a matrix is invertible. A non-zero determinant means a matrix is invertible.
The determinant can also determine properties like scalability of volume in geometry.

In the original exercise, understanding that det((A + B^T)^T) = det(A^T + B) was crucial. This helps in showing that the determinants of the matrices involved are equal, thanks to the properties of determinants and transposes discussed.

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