Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Matrix Operations
Let \( A = \begin{bmatrix} 8 & -9 \\ 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 9 & -2 \\ 3 & 1 \end{bmatrix} \). Complete parts a through d.
#### (a) Find \( 5A + 6B \).
First, we need to find the matrices \( 5A \) and \( 6B \).
Calculate \( 5A \):
\[
5A = 5 \begin{bmatrix} 8 & -9 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 5 \cdot 8 & 5 \cdot (-9) \\ 5 \cdot 2 & 5 \cdot 3 \end{bmatrix} = \begin{bmatrix} 40 & -45 \\ 10 & 15 \end{bmatrix}
\]
Calculate \( 6B \):
\[
6B = 6 \begin{bmatrix} 9 & -2 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 6 \cdot 9 & 6 \cdot (-2) \\ 6 \cdot 3 & 6 \cdot 1 \end{bmatrix} = \begin{bmatrix} 54 & -12 \\ 18 & 6 \end{bmatrix}
\]
Now, add the matrices \( 5A \) and \( 6B \):
\[
5A + 6B = \begin{bmatrix} 40 & -45 \\ 10 & 15 \end{bmatrix} + \begin{bmatrix} 54 & -12 \\ 18 & 6 \end{bmatrix} = \begin{bmatrix} 40 + 54 & -45 + (-12) \\ 10 + 18 & 15 + 6 \end{bmatrix} = \begin{bmatrix} 94 & -57 \\ 28 & 21 \end{bmatrix}
\]
So,
\[
5A + 6B = \begin{bmatrix} 94 & -57 \\ 28 & 21 \end{bmatrix}
\]
#### (b) Find \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06bbd155-c9d1-4bdc-8881-0cd893ac016a%2F482dadc9-072f-4814-aee2-dfd37dab64b1%2Faam85u_processed.png&w=3840&q=75)
Transcribed Image Text:### Matrix Operations
Let \( A = \begin{bmatrix} 8 & -9 \\ 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 9 & -2 \\ 3 & 1 \end{bmatrix} \). Complete parts a through d.
#### (a) Find \( 5A + 6B \).
First, we need to find the matrices \( 5A \) and \( 6B \).
Calculate \( 5A \):
\[
5A = 5 \begin{bmatrix} 8 & -9 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 5 \cdot 8 & 5 \cdot (-9) \\ 5 \cdot 2 & 5 \cdot 3 \end{bmatrix} = \begin{bmatrix} 40 & -45 \\ 10 & 15 \end{bmatrix}
\]
Calculate \( 6B \):
\[
6B = 6 \begin{bmatrix} 9 & -2 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 6 \cdot 9 & 6 \cdot (-2) \\ 6 \cdot 3 & 6 \cdot 1 \end{bmatrix} = \begin{bmatrix} 54 & -12 \\ 18 & 6 \end{bmatrix}
\]
Now, add the matrices \( 5A \) and \( 6B \):
\[
5A + 6B = \begin{bmatrix} 40 & -45 \\ 10 & 15 \end{bmatrix} + \begin{bmatrix} 54 & -12 \\ 18 & 6 \end{bmatrix} = \begin{bmatrix} 40 + 54 & -45 + (-12) \\ 10 + 18 & 15 + 6 \end{bmatrix} = \begin{bmatrix} 94 & -57 \\ 28 & 21 \end{bmatrix}
\]
So,
\[
5A + 6B = \begin{bmatrix} 94 & -57 \\ 28 & 21 \end{bmatrix}
\]
#### (b) Find \(
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