Problem 16 - A First Course in Continuum Mechanics

Question

For any two second-order tensors \(\boldsymbol{A}\) and \(\boldsymbol{B}\) show that $$ \operatorname{det}(\boldsymbol{A B})=(\operatorname{det} \boldsymbol{A})(\operatorname{det} \boldsymbol{B}) $$ Moreover, if \(\boldsymbol{A}^{-1}\) exists, show that $$ \operatorname{det} \boldsymbol{A}^{-1}=1 / \operatorname{det} \boldsymbol{A} $$

Step-by-Step Solution

Answer

\(\operatorname{det}(\boldsymbol{AB}) = \operatorname{det}(\boldsymbol{A}) \cdot \operatorname{det}(\boldsymbol{B})\) and \(\operatorname{det} \boldsymbol{A}^{-1} = 1 / \operatorname{det} \boldsymbol{A}\)

Step 1: Recall the property of determinants for product of two matrices

Remember that the determinant of a product of two matrices is equal to the product of their determinants. This is expressed as \( \operatorname{det}(\boldsymbol{AB}) = \operatorname{det}(\boldsymbol{A}) \cdot \operatorname{det}(\boldsymbol{B}) \) for any two square matrices \(\boldsymbol{A}\) and \(\boldsymbol{B}\) of the same size.

Step 2: Apply the determinant product property

Use the property directly to state that for second-order tensors (which can be treated as 2x2 matrices), \( \operatorname{det}(\boldsymbol{AB}) = \operatorname{det}(\boldsymbol{A}) \cdot \operatorname{det}(\boldsymbol{B}) \) holds true, confirming the first part of the exercise.

Step 3: Use the properties of inverse matrices and determinants

Knowing that \(\boldsymbol{A}^{-1} \) is the inverse of \( \boldsymbol{A} \) such that \( \boldsymbol{A} \boldsymbol{A}^{-1} = \boldsymbol{I} \) where \( \boldsymbol{I} \) is the identity matrix, and \( \operatorname{det}(\boldsymbol{I}) = 1 \) for any identity matrix, use these properties to show the second part of the exercise.

Step 4: Calculate the determinant of the identity matrix using the product of \( \boldsymbol{A} \) and its inverse

Since \( \boldsymbol{A} \boldsymbol{A}^{-1} = \boldsymbol{I} \) and applying the determinant product property, \( \operatorname{det}(\boldsymbol{A} \boldsymbol{A}^{-1}) = \operatorname{det}(\boldsymbol{A}) \cdot \operatorname{det}(\boldsymbol{A}^{-1}) = \operatorname{det}(\boldsymbol{I}) = 1 \).

Step 5: Solve for the determinant of the inverse

From the previous step, rearrange the equation to find \( \operatorname{det}(\boldsymbol{A}^{-1}) \) as follows: \( \operatorname{det}(\boldsymbol{A}) \cdot \operatorname{det}(\boldsymbol{A}^{-1}) = 1 \) implies that \( \operatorname{det}(\boldsymbol{A}^{-1}) = \frac{1}{\operatorname{det}(\boldsymbol{A})} \) thus proving the second part.

Key Concepts

Second-Order Tensors

When we speak about second-order tensors, we're delving into a type of mathematical object that is prevalent in various fields such as physics and engineering. These tensors are essentially generalized matrices with two indices that can represent numerous physical quantities like stress or strain in materials.

In the context of the exercise, we treat second-order tensors analogous to 2x2 square matrices, which allows us to apply matrix operations and properties to these tensors. It is critical to understand that the operation of taking a determinant is not just limited to matrices but extends to tensors as well, serving a similar purpose in revealing certain intrinsic properties related to the space they operate within, for example, the 'volume' transformation resulting from a linear transformation that the tensor might represent.

Furthermore, in the realm of tensors, just like matrices, operations such as multiplication and finding inverses are possible, which directly leads us to the exploration of their determinant properties. Comprehending the fundamental aspects of second-order tensors is key to mastering more complex concepts in linear algebra and tensor calculus.

Matrix Determinant Properties

The determinant is a scalar attribute of a square matrix that can tell us much about the matrix itself. It is often used to determine whether a matrix is invertible, with a non-zero determinant indicating an invertible matrix and a zero determinant signifying a singular, or non-invertible, matrix.

One of the essential properties used in the exercise is the product property of determinants, which states that the determinant of the product of two matrices is equal to the product of their determinants. This can be mathematically stated as \( \operatorname{det}(\boldsymbol{A} \boldsymbol{B}) = \operatorname{det}(\boldsymbol{A}) \cdot \operatorname{det}(\boldsymbol{B}) \).

This property is powerful because it simplifies the evaluation of the determinant of large matrices and provides a foundation for understanding how linear transformations (represented by matrices) combine. Another crucial aspect is that this property holds irrespective of the order of multiplication, a reflection of the commutative property of scalar multiplication.

Inverse Matrices

Inverse matrices hold a significant place in linear algebra because they provide a method to 'undo' the effects of a given linear transformation. For a square matrix \(\boldsymbol{A}\), its inverse \(\boldsymbol{A}^{-1}\) is a matrix that, when multiplied by \(\boldsymbol{A}\), yields the identity matrix \(\boldsymbol{I}\). The identity matrix is like a neutral element in matrix multiplication, akin to the number 1 in scalar multiplication.

As shown in the solution, the determinant of an inverse matrix is a reciprocal of the determinant of the original matrix. Mathematically, we express this as \(\operatorname{det}(\boldsymbol{A}^{-1}) = 1 / \operatorname{det}(\boldsymbol{A})\). This relationship plays a pivotal role in solving systems of linear equations and in understanding the geometric implications of matrix transformations, such as rotations and reflections.

It's important to realize that not all matrices have an inverse; only those with a non-zero determinant do. This leads us full circle back to the determinant's critical role in matrix theory, contributing to the foundation necessary for advanced mathematical applications.