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,...
Related questions
Question
![**Matrix Operations: Example Problem**
*Perform the indicated operation.*
**Problem:**
Let
\[ C = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]
and
\[ D = \begin{bmatrix} -1 \\ 3 \end{bmatrix} \]
Find \( C - 3D \).
**Solution:**
To solve the problem, calculate the expression \( C - 3D \).
1. **Calculate \( 3D \):**
Multiply each element of matrix \( D \) by 3:
\[ 3 \times \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 3(-1) \\ 3(3) \end{bmatrix} = \begin{bmatrix} -3 \\ 9 \end{bmatrix} \]
2. **Subtract \( 3D \) from \( C \):**
\[ C - 3D = \begin{bmatrix} 1 \\ 2 \end{bmatrix} - \begin{bmatrix} -3 \\ 9 \end{bmatrix} = \begin{bmatrix} 1 - (-3) \\ 2 - 9 \end{bmatrix} = \begin{bmatrix} 4 \\ -7 \end{bmatrix} \]
Therefore, the result is:
\[ C - 3D = \begin{bmatrix} 4 \\ -7 \end{bmatrix} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff73336ba-b7a2-41c3-9bb8-e711f85d25de%2F4adb12db-df48-44a6-b9a8-e6f6da5ed8b9%2F8y15prn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Matrix Operations: Example Problem**
*Perform the indicated operation.*
**Problem:**
Let
\[ C = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]
and
\[ D = \begin{bmatrix} -1 \\ 3 \end{bmatrix} \]
Find \( C - 3D \).
**Solution:**
To solve the problem, calculate the expression \( C - 3D \).
1. **Calculate \( 3D \):**
Multiply each element of matrix \( D \) by 3:
\[ 3 \times \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 3(-1) \\ 3(3) \end{bmatrix} = \begin{bmatrix} -3 \\ 9 \end{bmatrix} \]
2. **Subtract \( 3D \) from \( C \):**
\[ C - 3D = \begin{bmatrix} 1 \\ 2 \end{bmatrix} - \begin{bmatrix} -3 \\ 9 \end{bmatrix} = \begin{bmatrix} 1 - (-3) \\ 2 - 9 \end{bmatrix} = \begin{bmatrix} 4 \\ -7 \end{bmatrix} \]
Therefore, the result is:
\[ C - 3D = \begin{bmatrix} 4 \\ -7 \end{bmatrix} \]
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