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Show that (AB)¹ = BTA¹ if A = 2 0 2 0 and B = -1 1 15) -3 1 -1 -1 2 2). a

Show that (AB)¹ = BTA¹ if A = 2 0 2 0 and B = -1 1 15) -3 1 -1 -1 2 2). a

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Linear Algebra: Matrix Transposition Properties **Problem Statement:** Show that \((AB)^T = B^T A^T\) if: \[ A = \begin{pmatrix} 2 & -3 & 1 \\ 0 & -1 & 2 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 2 & 0 \\ -1 & 1 \\ 1 & 5 \end{pmatrix}. \] **Explanation:** To prove \((AB)^T = B^T A^T\), we will go through the following steps: 1. **Compute the product \(AB\):** \[ AB = \begin{pmatrix} 2 & -3 & 1 \\ 0 & -1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ -1 & 1 \\ 1 & 5 \end{pmatrix} \] 2. **Compute the transpose of the product \((AB)^T\):** Once we have \(AB\), we will transpose the resulting matrix. 3. **Compute \(B^T\) and \(A^T\):** \[ B^T = \begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 5 \end{pmatrix} \quad \text{and} \quad A^T = \begin{pmatrix} 2 & 0 \\ -3 & -1 \\ 1 & 2 \end{pmatrix} \] 4. **Compute the product \(B^T A^T\):** \[ B^T A^T = \begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 5 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ -3 & -1 \\ 1 & 2 \end{pmatrix} \] Compare the matrices obtained from steps 2 and 4 to conclude that \((AB)^T = B^T A^T\). ### Detailed Calculation: #### Step 1: Compute \(AB\) \[ AB = \begin{pmatrix} 2 & -3 & 1 \\ 0 & -1 & 2 \end{pmatrix} \begin{pmatrix}
Transcribed Image Text:### Linear Algebra: Matrix Transposition Properties **Problem Statement:** Show that \((AB)^T = B^T A^T\) if: \[ A = \begin{pmatrix} 2 & -3 & 1 \\ 0 & -1 & 2 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 2 & 0 \\ -1 & 1 \\ 1 & 5 \end{pmatrix}. \] **Explanation:** To prove \((AB)^T = B^T A^T\), we will go through the following steps: 1. **Compute the product \(AB\):** \[ AB = \begin{pmatrix} 2 & -3 & 1 \\ 0 & -1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ -1 & 1 \\ 1 & 5 \end{pmatrix} \] 2. **Compute the transpose of the product \((AB)^T\):** Once we have \(AB\), we will transpose the resulting matrix. 3. **Compute \(B^T\) and \(A^T\):** \[ B^T = \begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 5 \end{pmatrix} \quad \text{and} \quad A^T = \begin{pmatrix} 2 & 0 \\ -3 & -1 \\ 1 & 2 \end{pmatrix} \] 4. **Compute the product \(B^T A^T\):** \[ B^T A^T = \begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 5 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ -3 & -1 \\ 1 & 2 \end{pmatrix} \] Compare the matrices obtained from steps 2 and 4 to conclude that \((AB)^T = B^T A^T\). ### Detailed Calculation: #### Step 1: Compute \(AB\) \[ AB = \begin{pmatrix} 2 & -3 & 1 \\ 0 & -1 & 2 \end{pmatrix} \begin{pmatrix}
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