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Chapter 8: Problem 33

Use Gaussian elimination to solve the system of linear equations. If there is no solution, state that the system is inconsistent. $$\left\\{\begin{aligned} r+2 s &=1 \\ 3 r+5 s+4 t &=7 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \( r = -7 + 8t \) and \( s = 4 - 4t \) for any real number t.

Step by step solution

01

Form the Augmented Matrix

Represent the system of equations as an augmented matrix. The augmented matrix is \[\begin{bmatrix} 1 & 2 & 0 & | & 1 \\ 3 & 5 & 4 & | & 7 \end{bmatrix}\]
02

Apply Gaussian elimination

Swap row 1 and row 2 to have the row with the largest leftmost nonzero entry on top. Then subtract 3 times the first row from the second row. The matrix becomes \[\begin{bmatrix} 1 & 2 & 0 & | & 1 \\ 0 & -1 & 4 & | & 4 \end{bmatrix}\]
03

Normalize the matrix

Multiply the 2nd row by -1 to get the coefficient of s to be positive. We get \[\begin{bmatrix} 1 & 2 & 0 & | & 1 \\ 0 & 1 & -4 & | & -4 \end{bmatrix}\]
04

Use back substitution

From the 2nd row, we have: s = 4 - 4t. Substitute 4 - 4t for s in the first equation to solve for other variables. Solving it yields r = -7 + 8t.

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Key Concepts

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

Linear Equations
Linear equations are a fundamental aspect of mathematics used to represent relationships between quantities. In their simplest form, linear equations involve two variables and can be written as \(ax + by = c\). These equations form straight lines in a two-dimensional coordinate system.
A system of linear equations involves two or more linear equations that are considered simultaneously. These systems can be represented in standard or matrix form and are often solved for variables that satisfy all equations in the system. In our example, the linear equations given are:
  • \(r + 2s = 1\)
  • \(3r + 5s + 4t = 7\)
The goal is to find the values of \(r\), \(s\), and \(t\) that satisfy both equations.
Augmented Matrix
An augmented matrix is a compact way of representing a system of linear equations. It includes the coefficients of the variables and the constants from the equations as a matrix. This representation helps simplify calculations during the solving process.
In an augmented matrix, the vertical line separates the coefficient matrix from the constant matrix. For our case, the initial augmented matrix is:
\[\begin{bmatrix} 1 & 2 & 0 & | & 1 \ 3 & 5 & 4 & | & 7 \end{bmatrix} \]
Here, the numbers before the vertical line represent coefficients of \(r\), \(s\), and \(t\), while the numbers after the line are the constants of the equations.
Back Substitution
Back substitution is a method used to find the unknown variables in a system of equations after performing Gaussian elimination. Once the system is transformed into an upper triangular or reduced row-echelon form, where rows below the main diagonal contain only zeros, variables can be determined starting from the last row.
Let's consider our transformed matrix:
\[ \begin{bmatrix} 1 & 2 & 0 & | & 1 \ 0 & 1 & -4 & | & -4 \end{bmatrix} \]
  • From the second row, we have \(s = 4 - 4t\).
  • We can substitute this value back into the first row, resulting in \(r = -7 + 8t\).
Back substitution systematically solves the variables starting from the simplest equation.
System of Equations
A system of equations consists of multiple equations with common variables. Solving these systems involves finding the values of these variables that simultaneously satisfy all equations. There are several methods to solve systems of equations, including substitution, elimination, and matrix-based methods like Gaussian elimination.
Systems can be categorized as:
  • Consistent: If there is at least one solution.
  • Inconsistent: If there is no solution.
  • Dependent: If there are infinitely many solutions.
In our exercise, Gaussian elimination helped determine the solution for the system, demonstrating that this particular system is consistent, given definitive values for \(r\), \(s\), and \(t\) using back substitution.

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Most popular questions from this chapter

A chemist wishes to make 10 gallons of a \(15 \%\) acid solution by mixing a \(10 \%\) acid solution with a \(25 \%\) acid solution. (a) Let \(x\) and \(y\) denote the total volumes (in gallons) of the \(10 \%\) and \(25 \%\) solutions, respectively. Using the variables \(x\) and \(y,\) write an equation for the total volume of the \(15 \%\) solution (the mixture). (b) Using the variables \(x\) and \(y,\) write an equation for the total volume of acid in the mixture by noting that Volume of acid in \(15 \%\) solution \(=\) volume of acid in \(10 \%\) solution \(+\) volume of acid in \(25 \%\) solution. (c) Solve the system of equations from parts (a) and (b), and interpret your solution. (d) Is it possible to obtain a \(5 \%\) acid solution by mixing a \(10 \%\) solution with a \(25 \%\) solution? Explain without solving any equations.

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A telephone company manufactures two different models of phones: Model 120 is cordless and Model 140 is not cordless. It takes 1 hour to manufacture the cordless model and 1 hour and 30 minutes to manufacture the traditional phone. At least 300 of the cordless models are to be produced. The manufacturer realizes a profit per phone of \(\$ 12\) for Model 120 and \(\$ 10\) for Model \(140 .\) If at most 1000 hours are to be allocated to the manufacture of the two models combined, how many of each model should be produced to maximize the total profit?

Consider the following system of equations.$$\left\\{\begin{array}{r} x+y=3 \\\\-x+y=1 \\\2 x+y=6\end{array}\right.$$ Use Gauss-Jordan elimination to show that this system has no solution. Interpret your answer in terms of the graphs of the given equations.

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