/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Consider the system of equations... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 5

Consider the system of equations $$\begin{array}{cc} 4 x+8 y-4 z & =4 \\ 3 x+8 y+5 z & =-11 \\ -2 x+y+12 z & =-17 \end{array}$$ a Set up the augmented matrix for this system. b Use row reduction to find the solution.

Short Answer

Expert verified
Solution: \( x = -3 \), \( y = 1 \), \( z = -2 \).

Step by step solution

01

Write the Augmented Matrix

First, we convert the given system of equations into an augmented matrix. The augmented matrix for the system \[ \begin{aligned} 4x + 8y - 4z &= 4\ 3x + 8y + 5z &= -11\ -2x + y + 12z &= -17 \end{aligned} \] is: \[ \left[ \begin{array}{ccc|c} 4 & 8 & -4 & 4 \ 3 & 8 & 5 & -11 \ -2 & 1 & 12 & -17 \end{array} \right] \].
02

First Row Operation (Create Leading 1 in Row 1)

To simplify, we make the leading coefficient of row 1 a 1 by dividing the first row by 4: \[ \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 1 \ 3 & 8 & 5 & -11 \ -2 & 1 & 12 & -17 \end{array} \right] \]
03

Second Row Operation (Eliminate First Column Below Row 1)

Use row operations to make the elements below the leading 1 in the first column zero. Replace rows 2 and 3: - Row 2: R2 = R2 - 3*R1- Row 3: R3 = R3 + 2*R1 After these operations, the matrix becomes: \[ \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 1 \ 0 & 2 & 8 & -14 \ 0 & 5 & 10 & -15 \end{array} \right] \]
04

Third Row Operation (Make Leading Coefficient in Row 2 a 1)

Divide the second row by 2 to have a leading 1: \[ \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 1 \ 0 & 1 & 4 & -7 \ 0 & 5 & 10 & -15 \end{array} \right] \]
05

Fourth Row Operation (Eliminate Second Column in Row 3)

Substitute row 3 to eliminate the second column value using row 2. Modify row 3: R3 = R3 - 5*R2. The matrix now transforms into: \[ \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 1 \ 0 & 1 & 4 & -7 \ 0 & 0 & -10 & 20 \end{array} \right] \]
06

Fifth Row Operation (Make Leading Coefficient in Row 3 a 1)

Divide row 3 by -10 to solve for z:\[ \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 1 \ 0 & 1 & 4 & -7 \ 0 & 0 & 1 & -2 \end{array} \right] \]
07

Back Substitution

Use back substitution to solve for y and x. 1. From row 3: \( z = -2 \)2. Substitute \( z = -2 \) into row 2: \( y + 4(-2) = -7 \rightarrow y - 8 = -7 \rightarrow y = 1 \)3. Substitute \( y = 1 \) and \( z = -2 \) into row 1: \( x + 2(1) - 1(-2) = 1 \rightarrow x + 2 + 2 = 1 \rightarrow x = -3 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Augmented Matrix
An augmented matrix is a crucial tool when solving systems of linear equations. Essentially, it combines the coefficients and constants of a system into a matrix form. In this format, the matrix has a vertical line dividing the coefficients of the variables on the left and the constant terms on the right.

To create an augmented matrix from the system of equations:
  • The coefficients of the variables form the main body of the matrix.
  • A vertical line separates these coefficients from the constants, which are placed in the last column.

This representation is especially useful because it allows us to apply matrix operations systematically to solve the system. It simplifies complex systems into a more organized format, making it easier to use techniques such as row reduction and Gaussian elimination.
Row Reduction
Row reduction is a technique used to simplify matrices to find the solutions of a system of equations. The main goal is to manipulate the matrix to a simpler form, usually called the row echelon form or even further to the reduced row echelon form.

Here鈥檚 how we perform row reduction:
  • First, create a leading 1 in each row of the matrix by dividing or multiplying the rows.
  • Then, use these leading 1s to eliminate all the other numbers in the same column by performing row operations like adding or subtracting multiples of one row from another.

By performing these row operations, we systematically simplify the matrix while maintaining the same solutions for the system. Once we reach a form where variables can be easily computed, back substitution is used to find the specific values for each variable.
Gaussian Elimination
Gaussian elimination is a method used to solve systems of linear equations. It simplifies the matrices through row operations, aiming to bring it to a form where solutions can be read easily from the resulting rows.

The primary steps include:
  • Transform the matrix into an upper triangular form (where all elements below the main diagonal are zeros).
  • Arrive at a row echelon or reduced row echelon form.
  • Use back substitution to find the solutions for each variable.

One of the primary benefits of Gaussian elimination is its systematic approach, which ensures each step logically follows the previous one. This method is foundational in linear algebra and helps students grasp the mechanics behind solving complex systems of equations through matrices.
Systems of Linear Equations
Systems of linear equations consist of multiple linear equations involving multiple variables. They can represent various real-world scenarios, from economic models to engineering problems.

Solving these systems involves finding the values of the variables that satisfy all the equations simultaneously. There are a few possible outcomes when solving these systems:
  • A unique solution, where a single set of values satisfies all equations.
  • No solution, indicating the equations are contradictory and cannot all be satisfied at once.
  • Infinitely many solutions, usually occurring when equations are dependent and describe the same line or plane.

Understanding systems of linear equations is crucial, as they form the backbone of many mathematical applications and problem-solving techniques.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

We begin with a deck of cards which has \(50 \%\) aces (A) and \(50 \%\) kings (K). Aces rank higher than kings. Player 1 is dealt one card, face down. Player 1 can look at the card, but does not show the card to Player 2. Player 1 then says "ace" or "king" depending on what his card is. Player 1 can either tell the truth and say what the card is \((\mathrm{T}),\) or he can bluff and say that he has a higher ranking card (B). Note that if Player 1 has an ace, he must tell the truth since there are no higher ranking cards. However, if he is dealt a king, he can bluff, by saying he has an ace. If Player 1 says "king" the game ends and both players break even. If Player 1 says "ace" then Player 2 can either call (C) or fold (F). If Player 2 folds, then Player 1 wins $$\$ 0.50$$If Player 2 calls and Player 1 does not have an ace, then Player 2 wins $$\$ 1 .$$ If Player 2 calls and Player 1 does have an ace, then Player 1 wins $$\$ 1 . $$

Consider the system of equations $$\begin{array}{cc} 3 x+2 y-w-v & =0 \\ 2 x-y+3 z+w+5 v & =0 \\ x+2 y+6 z-w & =0 \\ -y+z-3 w+v & =0 \\ x+y+z+w+v & =1 \end{array}$$ a Set up the augmented matrix for this system. b Use row reduction to find the solution. Now we are ready to apply everything we have learned about solving repeated zero-sum games to a much more challenging game in the next section!

Solve the system of equations. $$\begin{array}{c} 4 p_{1}-2 p_{2}=0 \\ p_{1}+p_{2}=1 \end{array}$$ For larger matrices, you can edit the Sage cell by adding additional terms in each row, and adding more rows. For example, you can replace [3,-5,1],[1,1,3] with [4,8,-4,4],[3,8,5,-11],[-2,1,12,-17].

Pure strategy equilibria. Explain why the Matching Pennies game has no pure strategy equilibrium point.

Do you want to play in a predictable pattern or randomly? What are some advantages and disadvantages of a pattern? What are some advantages and disadvantages of a random strategy? Hopefully, you concluded that the best strategy against our perfect player would be to play ROCK, PAPER, SCISSORS \(1 / 3\) of the time each, and to play randomly. We can say that our strategy is to play each option randomly with a probability of \(1 / 3,\) and call this the Random \((1 / 3,1 / 3,1 / 3)\) strategy.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.