Question

Let \(A, B\), and \(C\) be \(n \times n\) matrices, with \(A\) and \(B\) invertible. Show that a. If \(A\) commutes with \(C,\) then \(A^{-1}\) commutes with C. b. If \(A\) commutes with \(B\), then \(A^{-1}\) commutes with \(B^{-1}\)

Short answer

If matrices commute, their inverses also commute.

Step-by-step solution

Understand Commutativity

Two matrices \(A\) and \(B\) are said to commute if their product is the same when taken in either order, i.e., \(AB = BA\). This is the key property we'll examine with the inverses of given matrices.

Apply Commutativity to Inverses for (a)

Given \(AC = CA\), we need to prove that \(A^{-1}C = CA^{-1}\). Start by multiplying both sides of the equation \(AC = CA\) by \(A^{-1}\) on the left: \[A^{-1}(AC) = A^{-1}(CA)\] This simplifies to: \[(A^{-1}A)C = (A^{-1}C)A\] Since \(A^{-1}A = I\) (the identity matrix), it reduces to: \[IC = A^{-1}CA\] Thus, \(C = A^{-1}CA\). Finally, we multiply both sides of \(C = A^{-1}CA\) by \(A^{-1}\) on the right: \[CA^{-1} = A^{-1}C\]. Hence, \(A^{-1}\) commutes with \(C\).

Apply Commutativity to Inverses for (b)

Given \(AB = BA\), we need to prove that \(A^{-1}B^{-1} = B^{-1}A^{-1}\). Start by taking the inverse of both sides of \(AB = BA\):\[(AB)^{-1} = (BA)^{-1}\]. Using the property \((XY)^{-1} = Y^{-1}X^{-1}\), we get: \[B^{-1}A^{-1} = A^{-1}B^{-1}\]. Thus, \(A^{-1}\) commutes with \(B^{-1}\).

Conclusion

We have shown through manipulation of the given equations and properties of inverses that if matrices commute, their inverses also commute following the inverse property and distribution laws of matrices. Therefore, both parts (a) and (b) are proven.

Key concepts

Commutativity of Matrices

Commutativity is a fascinating property in matrix algebra. It's less common in matrices than you might think, because unlike regular numbers, multiplying matrices does not always yield the same results when the order of multiplication is switched. For two matrices, say \( A \) and \( B \), to commute

• It means that \( AB = BA \).
• This property is crucial when dealing with inverses, as seen in the exercises.

For commutation to occur, both matrices involved should have a very particular structure or property. This concept becomes especially significant when we start working with the inverses of matrices. For example, if \( A \) commutes with \( C \), the exercise proves that \( A^{-1} \) will also commute with \( C \). Understanding commutativity can make dealing with complex matrix problems much easier, as it allows you to move terms around flexibly, aiding in solutions.

Properties of Inverse Matrices

The inverse of a matrix is much like the concept of reciprocal in number algebra. If you multiply a matrix by its inverse, the result is an identity matrix, denoted by \( I \), which acts much like multiplying by 1 in basic arithmetic. To find an inverse:

• Not every matrix has an inverse. Only square, non-singular matrices have inverses. A non-singular matrix has a non-zero determinant.
• The inverse of matrix \( A \) is denoted \( A^{-1} \), and we have the property \( AA^{-1} = A^{-1}A = I \).

In our exercise, these properties allow us to demonstrate that if two matrices \( A \) and \( B \) commute, then their inverses \( A^{-1} \) and \( B^{-1} \) will also commute. This reveals the elegant symmetry in matrix algebra resulting from the characteristics of inverses.

Matrix Algebra

Matrix algebra forms a backbone for various calculations in linear algebra. It deals with operations like addition, multiplication, and finding inverses of matrices. Here’s what to keep in mind:

• Matrix addition is straightforward, performed element-wise, and is commutative.
• Matrix multiplication is more complex. It involves taking rows of the first matrix and columns of the second. Multiplication is not generally commutative.
• Inverses play a critical role when solving matrix equations, much like fractions in solving simple algebraic equations.

In the realm of matrix algebra, understanding how inverses work allows for solving linear systems and handling transformations in higher-dimensional space. The exercise demonstrates how commutation and inverse properties can simplify and clarify these processes, contributing to a more thorough understanding of matrix operations.