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Let A = Find 4A. 4A = 0 5 -2 -8 -

Let A = Find 4A. 4A = 0 5 -2 -8 -

Calculus: Early Transcendentals
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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Let \( A = \begin{bmatrix} 0 & -2 \\ 5 & -8 \end{bmatrix} \). Find \( 4A \). --- To find \( 4A \), multiply each element of the matrix \( A \) by 4. \[ 4A = 4 \times \begin{bmatrix} 0 & -2 \\ 5 & -8 \end{bmatrix} = \begin{bmatrix} 4 \times 0 & 4 \times -2 \\ 4 \times 5 & 4 \times -8 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 20 & -32 \end{bmatrix} \] Thus, \( 4A = \begin{bmatrix} 0 & -8 \\ 20 & -32 \end{bmatrix} \). This exercise is presented in a linear format with a step-by-step guide to help you understand matrix scalar multiplication.
Transcribed Image Text:Let \( A = \begin{bmatrix} 0 & -2 \\ 5 & -8 \end{bmatrix} \). Find \( 4A \). --- To find \( 4A \), multiply each element of the matrix \( A \) by 4. \[ 4A = 4 \times \begin{bmatrix} 0 & -2 \\ 5 & -8 \end{bmatrix} = \begin{bmatrix} 4 \times 0 & 4 \times -2 \\ 4 \times 5 & 4 \times -8 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 20 & -32 \end{bmatrix} \] Thus, \( 4A = \begin{bmatrix} 0 & -8 \\ 20 & -32 \end{bmatrix} \). This exercise is presented in a linear format with a step-by-step guide to help you understand matrix scalar multiplication.
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