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Chapter 8: Problem 21

Use Gaussian elimination to solve the system of linear equations. If there is no solution, state that the system is inconsistent. $$\left\\{\begin{aligned} x+2 z &=0 \\ y+z &=-1 \\ x+8 y+4 z &=1 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is x = 3, y = 0.5, z = -1.5.

Step by step solution

01

Write the given system into an augmented matrix

An augmented matrix represents a system of linear equations. It contains all coefficients and solutions of the original equations and rewrites them into matrix form. This system of equations \\[\left\{\\begin{aligned} x+2 z &=0 \\ y+z &=-1 \\ x+8 y+4 z &=1 \\\end{aligned}\\right.\\] can be represented as the following augmented matrix: \\[\left[\\begin{array}{ccc|c} 1&0&2&0 \\ 0&1&1&-1 \\ 1&8&4&1 \\\end{array}\\right]\\]
02

Perform elementary row operations to simplify

Here is the foundation of Gaussian elimination: transforming the original matrix by using elementary row operations until it's in a form where we can solve it easily (row echelon form or reduced row echelon form). Subtract the first row from the third to get rid of x in the third equation, and obtain \\[\left[\\begin{array}{ccc|c} 1&0&2&0 \\ 0&1&1&-1 \\ 0&8&2&1 \\\end{array}\\right]\\]
03

Further simplify by row operations

Subtract 8 times the second row from the third row to get rid of y there. You'll obtain \\[\left[\\begin{array}{ccc|c} 1&0&2&0 \\ 0&1&1&-1 \\ 0&0&-6&9 \\\end{array}\\right]\\] Now we can backtrack to find the solutions.
04

Back-substitute to solve the system

From the last row: -6z = 9, so z = -1.5. Substitute z into the second equation to find y and into the first to find x. x = -2*(-1.5) = 3, and y = -1 - (-1.5) = 0.5.

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Key Concepts

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

System of Linear Equations
A system of linear equations consists of two or more linear equations involving the same set of variables. When we try to solve these systems, we're looking for values of the variables that satisfy all equations simultaneously. In the given exercise, the system is composed of three equations with three variables: \(x\), \(y\), and \(z\). Each equation provides specific conditions that these variables must meet simultaneously.

Solving such systems can reveal crucial insights, such as:
  • Unique Solutions: One set of variables satisfies all equations.
  • Infinite Solutions: More than one solution exists; often depicted by a parameterized line.
  • No Solution: No set of variables can satisfy all equations at once.
Gaussian elimination is a systematic method for finding solutions, if they exist, by transforming the system into a simpler form.
Augmented Matrix
An augmented matrix represents a concise way to handle multiple equations simultaneously without writing down the variables each time. It combines the coefficients of each variable and the constants from the right side of the equations into a single matrix. In our problem, we started with the following system:

\[\begin{aligned} x + 2z &= 0 \ y + z &= -1 \ x + 8y + 4z &= 1 \end{aligned}\]
Its augmented matrix is:
\[\begin{bmatrix} 1 & 0 & 2 & | & 0 \ 0 & 1 & 1 & | & -1 \ 1 & 8 & 4 & | & 1 \end{bmatrix}\]
The vertical bar separates the coefficients from the constants. This matrix provides a structured way to apply operations and guide us through Gaussian elimination.
Elementary Row Operations
Elementary row operations are the tools we use to manipulate matrices in Gaussian elimination. They aim to transform a given matrix into an easier form (often a row echelon form), allowing us to extract solutions:
  • Swapping Rows: Any two rows can be interchanged.
  • Multiplying a Row: Any row can be multiplied by a non-zero constant.
  • Adding Rows: A multiple of one row can be added to another row.
In this exercise, we performed several elementary row operations:
- First, we subtracted the first row from the third row to remove \(x\) from the third equation, simplifying our system.- Then, 8 times the second row was subtracted from the third, removing \(y\) from it. This brought us to a form where we could easily back-substitute to find \(z\), \(y\), and \(x\).

These operations maintain the original system's solution while moving us towards a clearer path to solve it.

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Most popular questions from this chapter

Find the decoding matrix for each encoding matrix. $$\left[\begin{array}{rrrr}1 & 1 & 4 & 1 \\\2 & -3 & 4 & 1 \\\3 & -4 & 6 & 2 \\\\-1 & 0 & -2 & -1\end{array}\right]$$

Adult and children's tickets for a certain show sell for \(\$ 8\) each. A total of 1000 tickets are sold, with total sales of \(\$ 8000 .\) Is it possible to figure out exactly how many of each type of ticket were sold? Why or why not?

A family owns and operates three businesses. On their income-tax return, they have to report the depreciation deductions for the three businesses separately. In \(2004,\) their depreciation deductions consisted of use of a car, plus depreciation on 5 -year equipment (on which onc-fifth of the original value is deductible per year) and 10-year equipment (on which one- tenth of the original valuc is deductible per year). The car use (in miles) for cach business in 2004 is given in the following table, along with the original value of the depreciable 5 - and 10 year equipment used in each business that year. $$\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { Car } \\\\\text { Use }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 5-Year }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 10-Year }\end{array} \\\\\text { Business } & \text { (miles) } & \text { Equipment (S) } & \text { Equipment (S) } \\\\\hline 1 & 3200 & 9850 & 435 \\\2 & 8800 & 12,730 & 980 \\\3 & 6880 & 2240 & 615\\\\\hline\end{array}$$ The depreciation deduction for car use in 2004 was 37.5 cents per mile. Use matrix multiplication to determine the total depreciation deduction for each business in 2004.

The volume of a paper party hat, shaped in the form of a right circular cone, is \(36 \pi\) cubic inches. If the radius of the cone is one-fourth the height of the cone, find the radius and the height.

A couple has \(\$ 10,000\) to invest for their child's wedding. Their accountant recommends placing at least \(\$ 6000\) in a high-yield investment and no more than \(\$ 4000\) in a low-yield investment. (a) Use \(x\) to denote the amount of money placed into the high-yield investment. Use \(y\) to denote the amount of money placed into the low-yield investment. Write a system of linear inequalities that describes the possible amounts the couple could invest in each type of venture. (b) Graph the region that represents all possible amounts the couple could put into each investment if they wish to follow the accountant's advice.
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