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Chapter 3: Problem 49

$$ \begin{array}{rlr} \text { a. }-x-4 y+2 z & =4 & \text { b. }-x-4 y+2 z=0 \\ x+2 y-z & =3 & x+2 y-z=3 \\ x+y-z & =8 & x+y-z=2 \\ \text { c. }-x-4 y+2 z & =0 & \\ x+2 y-z & =0 & \\ x+y-z & =0 & \end{array} $$

Short Answer

Expert verified
The short versions of the answers for each system are as follows: System A: \((x, y, z) = (6, 1, 5)\) System B: \(\left(x, y, z\right) = \left(-\frac{1}{3}, \frac{7}{3}, -\frac{5}{3}\right)\) System C: Infinitely many solutions (dependent system).

Step by step solution

01

Eliminate x from the first pair of equations (E1 and E2) in each system.

Add E1 and E2 in each of the systems. System A: E1 + E2: 2y + z = 7 System B: E1 + E2: 2y + z = 3 System C: E1 + E2: 2y + z = 0
02

Eliminate x from the first pair of equations (E1 and E3) in each system.

Add E1 and E3 in each of the systems. System A: E1 + E3: -3y + 3z = 12 System B: E1 + E3: -3y + 3z = 2 System C: E1 + E3: -3y + 3z = 0
03

Solve the resulting system in y and z for each system.

System A: \[ \begin{cases} 2y + z = 7 \\ -3y + 3z = 12 \end{cases} \] Multiply the first equation by 3 and add it to the second equation. 6y + 3z = 21 9y = 9 y = 1 Now, substitute the value of y into the first equation. 2(1) + z = 7 z = 5 Finally, substitute y and z's values into the original E2 of System A. x + 2(1) - 5 = 3 x = 6 System B: \[ \begin{cases} 2y + z = 3 \\ -3y + 3z = 2 \end{cases} \] Multiply the first equation by 3 and add it to the second equation. 6y + 3z = 9 3y = 7 y = \(\dfrac{7}{3}\) Now, substitute the value of y into the first equation. 2(\(\dfrac{7}{3}\)) + z = 3 z = -\(\dfrac{5}{3}\) Finally, substitute y and z's values into the original E2 of System B. x + 2(\(\dfrac{7}{3}\)) - (-\(\dfrac{5}{3}\)) = 3 x = -\(\dfrac{1}{3}\) System C: \[ \begin{cases} 2y + z = 0 \\ -3y + 3z = 0 \end{cases} \] The second equation represents a multiple of the first equation (-1.5 times), so there are infinitely many solutions. The system is dependent.
04

Write the solution for each system.

System A: \[ (x, y, z) = (6, 1, 5) \] System B: \[ (x, y, z) = \left(-\frac{1}{3}, \frac{7}{3}, -\frac{5}{3}\right) \] System C: Infinitely many solutions (dependent system).

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Key Concepts

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

System of Equations
A system of equations is a collection of two or more equations with a common set of unknowns. The challenge is to find values for these unknowns that make all equations true simultaneously.
In our examples, each system consists of three equations involving the variables \(x\), \(y\), and \(z\).
Here's a quick glimpse into the systems presented:
  • System A combines three equations, yielding one unique solution for \((x, y, z)\).
  • System B also contains three equations but results in a different, unique solution.
  • System C presents a situation with infinite solutions, highlighting the dependency of the equations on one another.
Understanding the behavior of these systems gives insight into how equations relate to each other and how they can be manipulated to find solutions.
Gaussian Elimination
Gaussian elimination is a straightforward method used in linear algebra to solve a system of linear equations.
It involves applying a sequence of operations to transform the system's augmented matrix into an upper triangular form, from which solutions can be deduced easily.
The main steps include:
  • Row Addition: Combine two equations to eliminate one of the variables. For instance, in both systems given, adding equations helped us get rid of \(x\).
  • Row Multiplication: Multiply an equation to facilitate elimination and scaling. We multiplied equations to line them up for easy addition and subtraction.
  • Back Substitution: Once the matrix is in upper triangular form, solve for one variable, and substitute back to find others. This is evident when resolving for \(y\) then \(z\), and finally \(x\) in our provided systems.
By strategically rearranging and simplifying the equations, Gaussian elimination simplifies complex systems into solvable formats.
Solution Sets
Solution sets describe the values that satisfy all equations in a system. They can take various forms:
  • Unique Solution: This occurs when there is only one set of values for the variables that solves the system. Systems A and B from our exercise are examples where unique solutions exist, resulting in precise values for \(x\), \(y\), and \(z\).
  • Infinite Solutions: When a system is dependent, like System C, there are countless solutions. This means the equations are essentially describing the same line or plane, leading to multiple overlapping answers.
  • No Solution: Although not present in our examples, this happens when the equations describe parallel lines or planes that never meet, indicating inconsistency and undetermined solutions.
Understanding how solution sets form and their implications allows us to interpret systems of equations better and recognize their broader mathematical impact.

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Most popular questions from this chapter

Compare addition and multiplication of \(1 \times 1\) matrices to the arithmetic of numbers.

Marketing Repeat Exercise 29 , given that the potential sales markets in the three cities are: Brakpan, 2,500 per day; Nigel, 1,500 per day; Springs, 1,200 per day.

Textbook Writing You are writing a college-level textbook on finite mathematics, and are trying to come up with the best combination of word problems. Over the years, you have accumulated a collection of amusing problems, serious applications, long complicated problems, and "generic" problems. \({ }^{25}\) Before your book is published, it must be scrutinized by several reviewers who, it seems, are never satisfied with the mix you use. You estimate that there are three kinds of reviewers: the "no-nonsense" types who prefer applications and generic problems, the "dead serious" types, who feel that a collegelevel text should be contain little or no humor and lots of long complicated problems, and the "laid-back" types, who believe that learning best takes place in a light-hearted atmosphere bordering on anarchy. You have drawn up the following chart, where the payoffs represent the reactions of reviewers on a scale of \(-10\) (ballistic) to \(+10\) (ecstatic): Reviewers ou \begin{tabular}{|l|c|c|c|} \hline & No-Nonsense & Dead Serious & Laid-Back \\ \hline Amusing & \(-5\) & \(-10\) & 10 \\ \hline Serious & 5 & 3 & 0 \\ \hline Long & \(-5\) & 5 & 3 \\ \hline Generic & 5 & 3 & \(-10\) \\ \hline \end{tabular} a. Your first draft of the book contained no generic problems, and equal numbers of the other categories. If half the reviewers of your book were "dead serious" and the rest were equally divided between the "no-nonsense" and "laid-back" types, what score would you expect? b. In your second draft of the book, you tried to balance the content by including some generic problems and eliminating several amusing ones, and wound up with a mix of which one eighth were amusing, one quarter were serious, three eighths were long, and a quarter were generic. What kind of reviewer would be least impressed by this mix? c. What kind of reviewer would be most impressed by the mix in your second draft?

Translate the given systems of equations into matrix form. \(x+y-z=8\) \(2 x+y+z=4\) \(\frac{3 x}{4}+\frac{z}{2}=1\)

Reduce the payoff matrices by dominance. $$ \begin{array}{r} \\\ a & b & c \\ 1 \\ \text { A } 2 \\ 3 & {\left[\begin{array}{rrr} 2 & -4 & -9 \\ -1 & -2 & -3 \\ 5 & 0 & -1 \end{array}\right]} \end{array} $$
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