/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Determine whether each of the fo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 1

Determine whether each of the following equations is linear: (a) \(5 x+7 y-8 y z=16\) (b) \(x+\pi y+e z=\log 5\) (c) \(3 x+k y-8 z=16\)

Short Answer

Expert verified
Equation (a) is nonlinear due to the product of variables in the term "-8yz". Equations (b) and (c) are linear, as all variables have a degree of 1 and there are no products of variables.

Step by step solution

01

Analyze equation (a)

For the equation (a) \(5x + 7y - 8yz = 16\), examine each term. We see that all the variables 'x' and 'y' have a degree of 1. However, in the term "-8yz", the variables 'y' and 'z' are multiplied together. This makes the equation nonlinear.
02

Analyze equation (b)

For the equation (b) \(x + \pi y + e z = \log 5\), examine each term. We see that all the variables 'x', 'y', and 'z' have a degree of 1. The constants \(\pi\), \(e\), and \(\log 5\) do not affect the linearity of the equation. Since there are no products of variables, equation (b) is linear.
03

Analyze equation (c)

For the equation (c) \(3x + ky - 8z = 16\), examine each term. We see that all the variables 'x', 'y', and 'z' have a degree of 1. The constant 'k' and '16' do not affect the linearity of the equation. Since there are no products of variables, equation (c) is linear. Therefore, equation (a) is nonlinear, and equations (b) and (c) are linear.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Reduce each of the following matrices to echelon form and then to row canonical form: (a) \(\left[\begin{array}{rrr}1 & 1 & 2 \\ 2 & 4 & 9 \\ 1 & 5 & 12\end{array}\right]\) (b) \(\left[\begin{array}{rrrrr}1 & 2 & -1 & 2 & 1 \\ 2 & 4 & 1 & -2 & 5 \\ 3 & 6 & 3 & -7 & 7\end{array}\right]\) (c) \(\left[\begin{array}{rrrrrr}2 & 4 & 2 & -2 & 5 & 1 \\ 3 & 6 & 2 & 2 & 0 & 4 \\ 4 & 8 & 2 & 6 & -5 & 7\end{array}\right]\)

Let \(A=\left[\begin{array}{rrrr}2 & -2 & 2 & 1 \\ -3 & 6 & 0 & -1 \\ 1 & -7 & 10 & 2\end{array}\right] .\) Reduce \(A\) to echelon form using the pivoting algorithm.

Describe the pivoting row-reduction algorithm. Also describe the advantages, if any, of using this pivoting algorithm.

Show that every elementary matrix \(E\) is invertible, and its inverse is an elementary matrix.

Use Gaussian elimination to solve the following systems of linear equations. \(x_{1}+2 x_{2}-x_{3}=-1\) (a) \(2 x_{1}+2 x_{2}+x_{3}=1\) (b) $\begin{aligned} x_{1}-2 x_{2}-x_{3} &=1 \\ 2 x_{1}-3 x_{2}+x_{3} &=6 \\\ 3 x_{1}-5 x_{2} &=7 \\ x_{1}+5 x_{3} &=9 \end{aligned}$ \(3 x_{1}+5 x_{2}-2 x_{3}=-1\) \(x_{1}+2 x_{2}+2 x_{4}=6\) (c) \(3 x_{1}+5 x_{2}-x_{3}+6 x_{4}=17\) \(2 x_{1}+4 x_{2}+x_{3}+2 x_{4}=12\) \(2 x_{1} \quad-7 x_{3}+11 x_{4}=7\) \(x_{1}-x_{2}-2 x_{3}+3 x_{4}=-63\) (d) \(2 x_{1}-x_{2}+6 x_{3}+6 x_{4}=-2\) \(-2 x_{1}+x_{2}-4 x_{3}-3 x_{4}=0\) \(3 x_{1}-2 x_{2}+9 x_{3}+10 x_{4}=-5\) \(x_{1}-4 x_{2}-x_{3}+x_{4}=3\) (f) \(x_{1}+2 x_{2}-x_{3}+3 x_{4}=2\) (e) \(2 x_{1}-8 x_{2}+x_{3}-4 x_{4}=9\) (f) \(2 x_{1}+4 x_{2}-x_{3}+6 x_{4}=5\) \(-x_{1}+4 x_{2}-2 x_{3}+5 x_{4}=-6\) \(x_{2}+2 x_{4}=3\) \(2 x_{1}-2 x_{2}-x_{3}+6 x_{4}-2 x_{5}=1\) (g) \(\quad x_{1}-x_{2}+x_{3}+2 x_{4}-x_{5}=2\) \(4 x_{1}-4 x_{2}+5 x_{3}+7 x_{4}-x_{5}=6\)Sec. \(3.4\) Systems of Linear Equations-Computational Aspects $$ \begin{aligned} 3 x_{1}-x_{2}+x_{3}-x_{4}+2 x_{5} &=5 \\ \text { (h) } x_{1}-x_{2}-x_{3}-2 x_{4}-x_{5} &=2 \\ 5 x_{1}-2 x_{2}+x_{3}-3 x_{4}+3 x_{5} &=10 \\ 2 x_{1}-x_{2}-2 x_{4}+x_{5} &=5 \\ 3 x_{1}-x_{2}+2 x_{3}+4 x_{4}+x_{5} &=2 \\ \text { (i) } x_{1}-x_{2}+2 x_{3}+3 x_{4}+x_{5} &=-1 \\ 2 x_{1}-3 x_{2}+6 x_{3}+9 x_{4}+4 x_{5} &=-5 \\ 7 x_{1}-2 x_{2}+4 x_{3}+8 x_{4}+x_{5} &=6 \\ 2 x_{1}+3 x_{3}-4 x_{5} &=5 \\ 3 x_{1}-4 x_{2}+8 x_{3}+3 x_{4} &=8 \\ x_{1}-x_{2}+2 x_{3}+x_{4}-x_{5} &=2 \\ -2 x_{1}+5 x_{2}-9 x_{3}-3 x_{4}-5 x_{5} &=-8 \end{aligned} $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.