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Chapter 8: Problem 85

Find two matrices \(A\) and \(B\) such that \(A B=B A\).

Short Answer

Expert verified
Matrices \(A\) and \(B\) where \(A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) and \(B= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\), commute as \(A B=B A\).

Step by step solution

01

Choose matrices A and B

A straightforward way to satisfy \(A B=B A\) is to choose either both \(A\) and \(B\) as the identity matrix, or to choose \(B\) as the identity matrix and \(A\) as any arbitrary matrix. For simplicity sake let's say \(A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) and \(B= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\), where \(B\) is the 2x2 identity matrix and \(A\) is a generic 2x2 matrix.
02

Perform matrix multiplication (AB and BA)

Multiply the matrices \(A\) and \(B\) in both ways. \n For \[ AB = A \cdot B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\] and \[ BA = B \cdot A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\] In both cases, we regain matrix \(A\), hence proving that both combinations of \(A\) and \(B\) result in the same product.
03

Verify result

Verify that \(A B=B A\). Both results from step2 are the same, hence, we have verified that \(A B=B A\) for the choice of \(A\) and \(B\) from step 1. This shows that the identity matrix commutes with all other matrices.

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Most popular questions from this chapter

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