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}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc156cfcf-b572-4dd1-9c92-c49112da643d%2F66b2a386-c98d-4e35-aa56-e156d873e983%2Fof4eljv_processed.png&w=3840&q=75)
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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