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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![### Matrix Addition and Scalar Multiplication
**Problem:**
Find \( 4A + 3B \).
Given matrices:
\[ A = \begin{bmatrix} 7 & 1 \\ 4 & 2 \end{bmatrix} \]
\[ B = \begin{bmatrix} -3 & 2 \\ 3 & -6 \end{bmatrix} \]
### Solution:
1. **Scalar Multiplication:**
First, multiply matrix \( A \) by 4:
\[
4A = 4 \times \begin{bmatrix} 7 & 1 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 4 \times 7 & 4 \times 1 \\ 4 \times 4 & 4 \times 2 \end{bmatrix} = \begin{bmatrix} 28 & 4 \\ 16 & 8 \end{bmatrix}
\]
Then, multiply matrix \( B \) by 3:
\[
3B = 3 \times \begin{bmatrix} -3 & 2 \\ 3 & -6 \end{bmatrix} = \begin{bmatrix} 3 \times -3 & 3 \times 2 \\ 3 \times 3 & 3 \times -6 \end{bmatrix} = \begin{bmatrix} -9 & 6 \\ 9 & -18 \end{bmatrix}
\]
2. **Matrix Addition:**
Now add the results from the above scalar multiplications:
\[
4A + 3B = \begin{bmatrix} 28 & 4 \\ 16 & 8 \end{bmatrix} + \begin{bmatrix} -9 & 6 \\ 9 & -18 \end{bmatrix} = \begin{bmatrix} 28 + (-9) & 4 + 6 \\ 16 + 9 & 8 + (-18) \end{bmatrix} = \begin{bmatrix} 19 & 10 \\ 25 & -10 \end{bmatrix}
\]
Therefore,
\[ 4A + 3B = \begin{bmatrix} 19 & 10 \\ 25 & -10 \end{bmatrix](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F006a35a3-e559-4cc9-aa53-746bb30a70e8%2Fc0f3b9ce-fbf4-4c03-af95-3cf417b2aa87%2Fve2lpq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Matrix Addition and Scalar Multiplication
**Problem:**
Find \( 4A + 3B \).
Given matrices:
\[ A = \begin{bmatrix} 7 & 1 \\ 4 & 2 \end{bmatrix} \]
\[ B = \begin{bmatrix} -3 & 2 \\ 3 & -6 \end{bmatrix} \]
### Solution:
1. **Scalar Multiplication:**
First, multiply matrix \( A \) by 4:
\[
4A = 4 \times \begin{bmatrix} 7 & 1 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 4 \times 7 & 4 \times 1 \\ 4 \times 4 & 4 \times 2 \end{bmatrix} = \begin{bmatrix} 28 & 4 \\ 16 & 8 \end{bmatrix}
\]
Then, multiply matrix \( B \) by 3:
\[
3B = 3 \times \begin{bmatrix} -3 & 2 \\ 3 & -6 \end{bmatrix} = \begin{bmatrix} 3 \times -3 & 3 \times 2 \\ 3 \times 3 & 3 \times -6 \end{bmatrix} = \begin{bmatrix} -9 & 6 \\ 9 & -18 \end{bmatrix}
\]
2. **Matrix Addition:**
Now add the results from the above scalar multiplications:
\[
4A + 3B = \begin{bmatrix} 28 & 4 \\ 16 & 8 \end{bmatrix} + \begin{bmatrix} -9 & 6 \\ 9 & -18 \end{bmatrix} = \begin{bmatrix} 28 + (-9) & 4 + 6 \\ 16 + 9 & 8 + (-18) \end{bmatrix} = \begin{bmatrix} 19 & 10 \\ 25 & -10 \end{bmatrix}
\]
Therefore,
\[ 4A + 3B = \begin{bmatrix} 19 & 10 \\ 25 & -10 \end{bmatrix
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