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Chapter 1: Problem 1

In each part, determine whether the equation is linear in \(x_{1}, x_{2},\) and \(x_{3}\) (a) \(x_{1}+5 x_{2}-\sqrt{2 x_{3}}=1\) (b) \(x_{1}+3 x_{2}+x_{1} x_{3}=2\) (c) \(x_{1}=-7 x_{2}+3 x_{3}\) (d) \(x_{1}^{-2}+x_{2}+8 x_{3}=5\) (e) \(x_{1}^{3 / 5}-2 x_{2}+x_{3}=4\) (f) \(\pi x_{1}-\sqrt{2 x_{2}}+\frac{1}{3} x_{3}=7^{1 / 3}\)

Short Answer

Expert verified
Only part (c) is linear in \(x_1, x_2,\) and \(x_3\). All other parts are non-linear.

Step by step solution

01

Understanding Linear Equations

A linear equation in variables \(x_1, x_2,\) and \(x_3\) is of the form \(a_1 x_1 + a_2 x_2 + a_3 x_3 = b\), where each \(a_i\) is a constant and each variable is raised only to the first power, meaning no products, roots or non-linear functions can be applied to \(x_1, x_2,\) or \(x_3\). Let's evaluate each equation against these criteria.
02

Evaluate Part (a)

The equation is \(x_1 + 5x_2 - \sqrt{2x_3} = 1\). The presence of \(\sqrt{2x_3}\) indicates a non-linear term due to the square root of \(x_3\). Hence, this equation is not linear in \(x_1, x_2,\) and \(x_3\).
03

Evaluate Part (b)

The equation is \(x_1 + 3x_2 + x_1 x_3 = 2\). The term \(x_1 x_3\) represents a product of variables, which is non-linear. Therefore, this equation is not linear in \(x_1, x_2,\) and \(x_3\).
04

Evaluate Part (c)

The equation is \(x_1 = -7x_2 + 3x_3\). All variables \(x_1, x_2,\) and \(x_3\) are to the first power and there are no products or roots. This equation is linear in \(x_1, x_2,\) and \(x_3\).
05

Evaluate Part (d)

The equation is \(x_1^{-2} + x_2 + 8x_3 = 5\). The term \(x_1^{-2}\) indicates an inverse squared, which is non-linear. Hence, this equation is not linear in \(x_1, x_2,\) and \(x_3\).
06

Evaluate Part (e)

The equation is \(x_1^{3/5} - 2x_2 + x_3 = 4\). The term \(x_1^{3/5}\) signifies a fractional power, implying non-linearity. Therefore, this equation is not linear in \(x_1, x_2,\) and \(x_3\).
07

Evaluate Part (f)

The equation is \(\pi x_1 - \sqrt{2x_2} + \frac{1}{3}x_3 = 7^{1/3}\). The term \(\sqrt{2x_2}\) involves a square root of a variable, making it non-linear. Hence, this equation is not linear in \(x_1, x_2,\) and \(x_3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-Linear Terms
In algebra, non-linear terms are any expressions in an equation that involve variables raised to a power other than one, involve roots, or are the product of two variables together. These terms disrupt the straight-line relationship that characterizes linear equations. For example, if you see a variable like \(x^2\) or \(\sqrt{x}\), you're dealing with non-linear terms. They can include:
  • Variables with exponents other than one: e.g., \(x^2\), \(x^{-3}\)
  • Square roots or other roots: e.g., \(\sqrt{x}\), \(\sqrt[3]{x}\)
  • Products of variables: e.g., \(xy\)
These elements make it impossible to graph the equation as a straight line when plotted.
First Power Variables
First power variables are essential to defining linear equations. When a variable is to the 'first power,' it simply means the variable is raised to an exponent of one, such as \(x^1\), which is normally written as just \(x\). In linear equations, every variable in the equation should be in its first power. Here are some characteristics:
  • No exponents other than one on any variable
  • No product of variables
  • No root or fractional powers
If you remember these guidelines, detecting an equation's linearity becomes straightforward: no fancy exponents, complexities, or cross-multiplications, just clean and simple algebra at play.
Linear Functions
A linear function represents a straight-line graph, typically expressed in the form \(y = mx + b\), where \(y\) is the dependent variable, \(m\) is the slope of the line, \(x\) is the independent variable, and \(b\) is the y-intercept. When dealing with equations, recognizing a linear function involves checking for the potential of a straight-line solution:
  • Each term either involves a single variable raised to the first power
  • No interaction between variables such as multiplication, division, or roots
  • A constant term \(b\) shifted vertically based on the value of the function
Linear functions are foundational as they form the basis of linear equations, which are prevalent across all levels of math and essential for solving real-world problems.
Sum of Terms
The sum of terms in an equation must respect the properties of linearity to be considered a linear equation. This means the equation is a collection (sum) of terms that each fit the criteria for a single linear term. For instance, the equation \(a_1x_1 + a_2x_2 + a_3x_3 = b\) is a sum of terms where:
  • \(a_i\) are constants or coefficients that scale each variable
  • Each \(x_i\) represents a variable appearing in the first power
  • The total equation balances on both sides with a constant \(b\)
When broken down, linear equations rely on the sum of such relationships to build a predictable and manageable form of mathematical expression, allowing an easy representation in graphs and real-world phenomena.

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Most popular questions from this chapter

Solve the following systems, where \(a, b,\) and \(c\) are constants. $$\begin{aligned}x_{1}+x_{2}+x_{3} &=a \\\2 x_{1}\quad+2 x_{3} &=b \\\3 x_{2}+3 x_{3} &=c \end{aligned}$$

Let \(A=\left[a_{i j}\right]\) be an \(n \times n\) matrix. Determine whether \(A\) is symmetric. (a) \(a_{i j}=i^{2}+j^{2}\) (b) \(a_{i j}=i^{2}-j^{2}\) (c) \(a_{i j}=2 i+2 j\) (d) \(a_{i j}=2 i^{2}+2 j^{3}\)

Solve the given homogeneous linear system by any method. $$\begin{aligned}2 x_{1}+x_{2}+3 x_{3} &=0 \\\x_{1}+2 x_{2} &=0 \\\x_{2}+x_{3} &=0 \end{aligned}$$

Let \(I\) be the \(n \times n\) matrix whose entry in row \(i\) and column \(j\) is $$\left\\{\begin{array}{ll} 1 & \text { if } i=j \\ 0 & \text { if } i \neq j \end{array}\right.$$ Show that \(A I=I A=A\) for every \(n \times n\) matrix \(A.\)

Determine the values of \(a\) for which the system has no solutions, exactly one solution, or infinitely many solutions. $$\begin{aligned}x+2 y-& \quad 3 z=\quad4 \\\3 x-y+\quad& 5 z=\quad 2\\\4 x+y+\left(a^{2}-14\right) z &=a+2 \end{aligned}$$
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