Question

Let \(A\) and \(B\) denote invertible \(n \times n\) matrices. Show that: a. \(\operatorname{adj}(\operatorname{adj} A)=(\operatorname{det} A)^{n-2} A\) (here \(\left.n \geq 2\right)\) [Hint: See Example 3.2.8.] b. \(\operatorname{adj}\left(A^{-1}\right)=(\operatorname{adj} A)^{-1}\) c. \(\operatorname{adj}\left(A^{T}\right)=(\operatorname{adj} A)^{T}\) d. \(\operatorname{adj}(A B)=(\operatorname{adj} B)(\operatorname{adj} A)\) [Hint: Show that \(A B \operatorname{adj}(A B)=A B \operatorname{adj} B \operatorname{adj} A .]\)

Short answer

Adjugates preserve det and transpose symmetries across matrix operations.

Step-by-step solution

Understanding Adjunct of a Matrix

The adjunct, or adjugate, of a matrix \( A \), denoted \( \operatorname{adj}(A) \), is related to the inverse of \( A \) as follows: for an \( n \times n \) matrix, \( A \operatorname{adj}(A) = \operatorname{adj}(A) A = (\det A) I \), where \( I \) is the identity matrix and \( \det A \) is the determinant of \( A \). Understanding this relationship will help prove each part of the problem.

Proving Part (a)

To show that \( \operatorname{adj}(\operatorname{adj} A)=(\det A)^{n-2} A \):1. Since \( \operatorname{adj}(A) \) is defined such that \( A \operatorname{adj}(A) = (\det A) I \), take \( \operatorname{adj} A \) and apply the adjugate operation again.2. Use the identity: \( A \operatorname{adj}(A \operatorname{adj}(A)) = ((\det A)^{n-1}) A \), which simplifies to \( (\det A)^{n-2} A \) when the adjugate of the adjugate is taken.3. Therefore, \( \operatorname{adj}(\operatorname{adj} A)=(\det A)^{n-2} A \).

Proving Part (b)

To prove \( \operatorname{adj}(A^{-1}) = (\operatorname{adj} A)^{-1} \):1. Recall that for any invertible matrix \( A \), \( A^{-1} = \frac{1}{\det A} \operatorname{adj}(A) \).2. For the inverses, it holds that \( \operatorname{adj}(A^{-1}) = \det(A) A^{n-2} \).3. Since \( (\operatorname{adj} A) A^{-1} = I \), their product is the identity matrix, which confirms the inverse relationship as: \( \operatorname{adj}(A^{-1}) = (\operatorname{adj} A)^{-1} \).

Proving Part (c)

To show \( \operatorname{adj}(A^T) = (\operatorname{adj} A)^T \):1. For any matrix \( A \), the identity \( (A^T)^{-1} = (A^{-1})^T \) holds.2. The adjugate of a transposed matrix relates as \( \operatorname{adj}(A^T) = \operatorname{adj}(A)^T \), due to the construction of cofactors and determinants, which are invariant under transposition.3. The equality follows by noting transposition does not change the laplacian matrix cofactors, thus \( \operatorname{adj}(A^T) = (\operatorname{adj} A)^T \).

Proving Part (d)

To show \( \operatorname{adj}(AB) = (\operatorname{adj} B)(\operatorname{adj} A) \):1. Expand using the properties \( AB \operatorname{adj}(AB) = (\det(AB)) I = (\det A)(\det B) I \).2. From this, \( A B \operatorname{adj} B \operatorname{adj} A = (\det A)(\det B) I \) holds, implying the adjugate is distributive over the multiplication in reverse order.3. Consequently, \( \operatorname{adj}(AB) = \operatorname{adj}(B) \times \operatorname{adj}(A) \).

Key concepts

Invertible Matrices

An invertible matrix, also called a non-singular or non-degenerate matrix, is a square matrix with a unique property: there exists another matrix, known as the inverse, which when multiplied with the original matrix results in the identity matrix. This property is crucial in linear algebra as it allows solving systems of linear equations, among other applications.

To determine if a matrix is invertible, check if its determinant \( \det A \) is not equal to zero. If the determinant is zero, the matrix doesn’t have an inverse.

• The concept of inverse can be expressed as \( A^{-1} \), and satisfies \( AA^{-1} = A^{-1}A = I \), where \( I \) is the identity matrix.
• Invertible matrices have full rank, meaning for an \( n \times n \) matrix, the rank is \( n \).

Understanding invertible matrices is foundational for exploring concepts like the adjugate matrix, described next.

Adjugate Matrix

The adjugate matrix, often denoted as \( \operatorname{adj}(A) \), is closely linked to both the determinant and the inverse of a matrix. It is constructed from the cofactors of the matrix and is the transpose of the cofactor matrix.

Here’s how it fits in matrix theory:

• The adjugate helps find the inverse of a matrix by the formula \( A^{-1} = \frac{1}{\det A} \operatorname{adj}(A) \).
• The relationship \( A \operatorname{adj}(A) = (\det A) I \) holds, where \( I \) is the identity matrix. This equation highlights the role of the adjugate in scaling the determinant to form inverses.

Adjugate matrices are essential for proofs involving transformations, such as those detailed in the exercise.

Matrix Determinant

The determinant is a special number associated with square matrices, serving as a key factor in matrix theory. It is depicted as \( \det A \) for a matrix \( A \). Determinants help to understand matrix functions and transformations, revealing whether a matrix is invertible.

Key properties include:

• If \( \det A = 0 \), the matrix is singular and not invertible.
• For an invertible matrix, \( \det A eq 0 \), ensuring the matrix has an inverse.
• Determinants provide scale factors for transformation, affecting areas and volumes when combined with matrices in geometric transformations.

In the exercise, determinants play a crucial role in proofs, particularly through identities that express relationships between matrices and their adjugates.

Matrix Transpose

The transpose of a matrix, notated as \( A^T \), is formed by flipping a matrix over its diagonal, converting rows into columns and vice versa. This simple yet powerful operation has numerous applications in matrix theory.

Understanding matrix transposition is key because:

• The transpose operation preserves certain properties like determinants and rank.
• In many equations, such as those involving dot products or solving linear systems, transposition makes it easier to handle matrix equations.
• Importantly, for the adjugate, \( \operatorname{adj}(A^T) = (\operatorname{adj}(A))^T \), showing how adjugates function with transposition.

Comprehending transposition and how it interacts with other matrix operations is essential for more advanced matrix theories as demonstrated in the exercise.