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Chapter 8: Problem 64

Consider the following system of equations.$$\left\\{\begin{array}{r}x+y=3 \\\\-x+y=1 \\\2 x+y=4\end{array}\right.$$ Use Gauss-Jordan elimination to find the solution, if it exists. Interpret your answer in terms of the graphs of the given equations.

Short Answer

Expert verified
The solution to the system of equations is \(x=0\) and \(y=1\). It represents the point of intersection of the three lines represented by the original equations in a graph.

Step by step solution

01

Write the Augmented Matrix of the System

Start by transforming the system of equations into an augmented matrix form. This means placing coefficients of \(x\), \(y\), and constants into a matrix in the order they occur: \[\begin{bmatrix}1 & 1 & 3 \\ -1 & 1 & 1 \\ 2 & 1 & 4\end{bmatrix}\] Each row here represents a separate equation.
02

Apply Row Operations

Begin elementary row operations for Gauss-Jordan elimination. This starts by transforming the matrix to row-echelon form by making elements below and above the leading entry (diagonal from top left to bottom right) to become zero. The operations could be: Swap row 1 and row 2, then add row 1 to row 2, and finally subtract two times row 1 from row 3. The new matrix will be: \[\begin{bmatrix}1 & -1 & -1 \\ 0 & 2 & 2 \\ 0 & 3 & 3\end{bmatrix}\]
03

Continue Row Operations

To proceed with the elimination, focus on making a leading 1 in each row. Begin by dividing row 2 by 2: \[\begin{bmatrix}1 & -1 & -1 \\ 0 & 1 & 1 \\ 0 & 3 & 3\end{bmatrix}\] Then subtract 3 times row 2 from row 3 to get: \[\begin{bmatrix}1 & -1 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}\] This operation will give a simplified form.
04

Convert Back to Equations

The current matrix can be converted back to equations to give the solution: \[x-y=-1\] and \[y=1\]. Substitute \(y = 1\) into the first equation to get \(x=0\). The solution to the system is therefore \(x=0\), \(y=1\).
05

Interpretation

Each of the original equations represents a straight line on a 2-D graph. The solution to the system (0,1) is the point where all three lines intersect. This is the graphical interpretation of the solution.

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Key Concepts

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

Understanding Systems of Equations
A system of equations is a set of two or more equations with the same variables. Solving a system means finding a set of values for the variables that satisfy all the equations simultaneously. For instance, in our system:
  • \(x + y = 3\)
  • \(-x + y = 1\)
  • \(2x + y = 4\)
We are looking for values of \(x\) and \(y\) that make all three equations true at once. Often, systems of equations arise in real-world problems and need consistent solutions.
If every equation is not satisfied with a single set of variables, the system may have either no solution or multiple solutions.
What is an Augmented Matrix?
An augmented matrix is a compact way to represent a system of linear equations. We take the coefficients of the variables and the constants from the right side of the equations and place them in one matrix. For our exercises, the augmented matrix looks like this:
  • \[\begin{bmatrix}1 & 1 & 3 \ -1 & 1 & 1 \ 2 & 1 & 4\end{bmatrix} \]
Each row corresponds to an equation, and each column corresponds to a variable or the constant. This format is particularly useful because it sets the stage for applying row operations to find a solution.
By handling our equations in matrix form, we can systematically simplify them using various techniques like Gauss-Jordan elimination.
Mastering Row Operations
Row operations are techniques used to manipulate the rows of an augmented matrix to achieve a simpler form. The main operations include:
  • Swapping two rows.
  • Multiplying a row by a non-zero scalar.
  • Adding or subtracting a multiple of one row to another.
In the Gauss-Jordan elimination method, the goal is to transform the matrix such that every variable has a leading 1, and other numbers in the variable's column become zeros. This leads to a matrix in row-echelon form or even reduced row-echelon form.
Continuing this way helps us isolate each variable and find its exact value. Here, we used row operations to arrive at a solution, showing that \(x = 0\) and \(y = 1\).
Graphical Interpretation of Solutions
Graphical interpretation helps us visualize the solution of a system as the point where the graphs of the equations intersect. In two dimensions, each linear equation in variables \(x\) and \(y\) can be represented as a straight line.
  • For our equations, the intersection point was found to be \((0, 1)\), which is where all three lines would cross on a graph.
This means that \((0,1)\) is the unique solution that satisfies all the equations simultaneously. Sometimes the graph can reveal specific characteristics of the system, such as parallel lines indicating no solution or overlapping lines indicating infinitely many solutions.
Hence, graphical representation is a powerful tool for understanding the nature of solutions to a system of equations.

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Most popular questions from this chapter

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.

According to health professionals, the daily intake of fat in a diet that consists of 2000 calories per day should not exceed 50 grams. The total fat content of a meal that consists of a Whopper and a medium order of fries exceeds this limit by 14 grams. Two Whoppers and a medium order of fries have a total fat content of 111 grams. Set up and solve a system of equations to find the fat content of a Whopper and the fat content of a medium order of fries.

Three students take courses at two different colleges, Woosamotta University \((\mathrm{WU})\) and Frostbite Falls Community College (FFCC). WU charges \(\$ 200\) per credit hour and FFCC charges \(\$ 120\) per credit hour. The number of credits taken by each student at each college is given in the following table. $$\begin{array}{|c|c|c|}\hline & {2}{c}\text { Credits } \\\\\text { Student } & \text { WU } & \text { FFCC } \\\\\hline 1 & 12 & 6 \\\2 & 3 & 9 \\\3 & 8 & 8 \\ \hline\end{array}$$ Use matrix multiplication to find the total tuition paid by cach student.

If \(A=\left[\begin{array}{rrr}1 & 0 & 1 \\ 0 & 0 & 1 \\ 2 & -1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{rrr}0 & 3 & -1 \\ -1 & 2 & 0 \\\ 0 & 0 & 1\end{array}\right],\) for what values of \(a\) and \(b\) does \(A B=\left[\begin{array}{rrr}0 & 2 a+2 b+1 & 0 \\ 3 a+4 b & 0 & 1 \\ 1 & 4 & -2\end{array}\right] ?\)

The following table lists the caloric content of a typical fast-food meal. Food (single serving) Calories Cheeseburger Medium order of fries Medium cola (21 oz) \(\begin{array}{lc}\text { Food (single serving) } & \text { Calories } \\\ \text { Cheeseburger } & 330 \\ \text { Medium order of fries } & 450 \\\ \text { Medium cola }(210 z) & 220\end{array}\) (a) After a lunch that consists of a cheeseburger, a medium order of fries, and a medium cola, you decide to burn off a quarter of the total calories in the meal by some combination of running and walking. You know that running burns 8 calories per minute and walking burns 3 calories per minute. If you exercise for a total of 40 minutes, how many minutes should you spend on each activity? (b) Rework part (a) for the case in which you exercise for a total of only 20 minutes. Do you get a realistic solution? Explain your answer.
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