/*! 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 2 Solve the following linear syste... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the following linear system: $$\begin{array}{l}2 x_{1}+5 x_{2}-8 x_{3}+2 x_{4}+2 x_{5}=0 \\ 6 x_{1}+2 x_{2}-10 x_{3}+6 x_{4}+8 x_{5}=6 \\ 3 x_{1}+6 x_{2}+2 x_{3}+3 x_{4}+5 x_{5}=6 \\ 3 x_{1}+1 x_{2}-5 x_{3}+3 x_{4}+4 x_{5}=3 \\ 6 x_{1}+7 x_{2}-3 x_{3}+6 x_{4}+9 x_{5}=9\end{array}$$ Be sure to set your work out carefully with equivalence signs \(\sim\) between each step, labeled by the row operations you performed.

Short Answer

Expert verified
No solution. The system is inconsistent.

Step by step solution

01

Form the Augmented Matrix

Write the augmented matrix from the given system of equations:\[\begin{bmatrix}2 & 5 & -8 & 2 & 2 & \vert & 0 \6 & 2 & -10 & 6 & 8 & \vert & 6 \3 & 6 & 2 & 3 & 5 & \vert & 6 \3 & 1 & -5 & 3 & 4 & \vert & 3 \6 & 7 & -3 & 6 & 9 & \vert & 9\end{bmatrix}\]
02

Row Transformations - Make First Pivot

Perform Row: \(R_2 \leftarrow R_2 - 3R_1\):\[\begin{bmatrix}2 & 5 & -8 & 2 & 2 & \vert & 0 \0 & -13 & 14 & 0 & 2 & \vert & 6 \3 & 6 & 2 & 3 & 5 & \vert & 6 \3 & 1 & -5 & 3 & 4 & \vert & 3 \6 & 7 & -3 & 6 & 9 & \vert & 9\end{bmatrix}\]
03

Additional Row Transformations

Perform Row: \(R_3 \leftarrow R_3 - \frac{3}{2}R_1\), \(R_4 \leftarrow R_4 - \frac{3}{2}R_1\), and \(R_5 \leftarrow R_5 - 3R_1\):\[\begin{bmatrix}2 & 5 & -8 & 2 & 2 & \vert & 0 \0 & -13 & 14 & 0 & 2 & \vert & 6 \0 & \frac{7}{2} & 14 & 0 & 2 & \vert & 6 \0 & -\frac{13}{2} & 7 & 0 & 1 & \vert & 3 \0 & -8 & 21 & 0 & 3 & \vert & 9\end{bmatrix}\]
04

Simplify Second Column

Perform Row: \(R_3 \leftarrow R_3 - \frac{7}{26}R_2\), \(R_4 \leftarrow R_4 + \frac{13}{26}R_2\), and \(R_5 \leftarrow R_5 + \frac{8}{13}R_2\):\[\begin{bmatrix}2 & 5 & -8 & 2 & 2 & \vert & 0 \0 & -13 & 14 & 0 & 2 & \vert & 6 \0 & 0 & 14.5 & 0 & 1.46 & \vert & 6 \0 & 0 & 0 & 0 & 0 & \vert & 4.64 \0 & 0 & 24.76 & 0 & 4.23 & \vert & 10.71\end{bmatrix}\]
05

Check for Consistency

Notice that Row 4 yields the equation \(0 = 4.64\), which is contradictory. This means the system of equations has no solution.

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 91Ó°ÊÓ!

Key Concepts

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

Augmented Matrix
When you're solving a system of linear equations, a powerful tool to use is the augmented matrix. This matrix takes the coefficients of the variables and the constants from the equations and presents them in a compact, organized manner. Each row of the matrix represents a single equation from the system, while each column represents the coefficients of one variable across all equations. A column at the end is separated by a vertical bar, which holds the constants from the right side of the equations. This setup helps streamline calculations and visualize the system as a whole. The augmented matrix for the system is written as follows, with each equation being a row and each variable coefficient being an entry in the matrix.
  • First column: coefficients of the first variable
  • Second column: coefficients of the second variable
  • Last column: constants from the equations
This forms the complete matrix that represents and aids in solving the system.
Row Operations
Row operations are the essential maneuvers used to simplify the augmented matrix into a more manageable form. The aim is often to reach a row-echelon form or reduced row-echelon form. These operations include:
  • Swapping two rows
  • Multiplying a row by a non-zero scalar
  • Adding or subtracting the multiple of one row from another row
Using these operations strategically allows you to move towards isolating the variables. By performing these systematic changes, you help illuminate the path towards the solution set or expose any issues like inconsistencies. As you progress through these steps, you write transformations such as: \( R_2 \leftarrow R_2 - 3R_1 \). Each step involves careful calculation to ensure accurate simplification of equations.
Inconsistency in Systems
A system of equations may occasionally exhibit inconsistencies, which means there is no solution that satisfies all the equations simultaneously. In the process of solving a system using matrices, an inconsistency might become apparent when you encounter a row where all variable coefficients are zero, but the constant on the right side isn't zero, translating into an equation like \(0 = c\), where \(c\) is a non-zero number. This indicates a contradiction, highlighting that the system doesn't have a solution. Such inconsistencies arise due to conflicting information within the equations and make it impossible to find values for the variables that work for all given equations.
Gaussian Elimination
Gaussian elimination is a systematic method used to solve systems of linear equations by transforming the system's augmented matrix into a simpler form, specifically an upper triangular matrix. This process involves using row operations to create zeros below the diagonal of the matrix. The ultimate goal is to deconstruct the system into simpler equations through back substitution, enabling you to solve for each variable step-by-step.Steps of Gaussian Elimination:
  • Start with the original augmented matrix of the system.
  • Use row operations to create zeros below the diagonal - this includes operations like \(R_2 \leftarrow R_2 - 3R_1\).
  • Once in triangular form, solve for the last variable first and build up until you get to the first variable, through back substitution.
Gaussian elimination is a robust technique for identifying solutions or detecting inconsistencies in a system, effectively managing complex interrelated equations to determine the outcome.

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

Another method for solving linear systems is to use row operations to bring the augmented matrix to Row Echelon Form (REF as opposed to RREF). In REF, the pivots are not necessarily set to one, and we only require that all entries left of the pivots are zero, not necessarily entries above a pivot. Provide a counterexample to show that row echelon form is not unique. Once a system is in row echelon form, it can be solved by "back substitution." Write the following row echelon matrix as a system of equations, then solve the system using back-substitution. $$\left(\begin{array}{lll|l}2 & 3 & 1 & 6 \\\0 & 1 & 1 & 2 \\\0 & 0 & 3 & 3\end{array}\right)$$

Write down examples of augmented matrices corresponding to each of the five types of solution sets for systems of equations with three unknowns.

State whether the following augmented matrices are in RREF and compute their solution sets. $$\begin{array}{l}\left(\begin{array}{lllll|l}1 & 0 & 0 & 0 & 3 & 1 \\ 0 & 1 & 0 & 0 & 1 & 2 \\\0 & 0 & 1 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 & 2 & 0\end{array}\right) \\ \left(\begin{array}{llllll|l}1 & 1 & 0 & 1 & 0 & 1 & 0 \\\0 & 0 & 1 & 2 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 & 3 & 0 \\\0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right),\end{array}$$ \(\left(\begin{array}{lllllll|r}1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 & 0 & 2 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\end{array}\right) .\)

Show that the RREF of a matrix is unique.

Is \(L U\) factorization of a matrix unique? Justify your answer. If you randomly create a matrix by picking numbers out of the blue, it will probably be difficult to perform elimination or factorization; fractions and large numbers will probably be involved. To invent simple problems it is better to start with a simple answer: (a) Start with any augmented matrix in RREF. Perform EROs to make most of the components non-zero. Write the result on a separate piece of paper and give it to your friend. Ask that friend to find RREF of the augmented matrix you gave them. Make sure they get the same augmented matrix you started with. (b) Create an upper triangular matrix \(U\) and a lower triangular matrix \(L\) with only 1s on the diagonal. Give the result to a friend to factor into \(L U\) form. (c) Do the same with an \(L D U\) factorization.

See all solutions

Recommended explanations on Math Textbooks

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.