/*! 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 34 Solve each system by Gaussian el... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each system by Gaussian elimination. $$ \begin{array}{l} -\frac{1}{3} x-\frac{1}{8} y+\frac{1}{6} z=-\frac{4}{3} \\ -\frac{2}{3} x-\frac{7}{8} y+\frac{1}{3} z=-\frac{23}{3} \\ -\frac{1}{3} x-\frac{5}{8} y+\frac{5}{6} z=0 \end{array} $$

Short Answer

Expert verified
The solution is \(x = 2\), \(y = 6\), \(z = -4\).

Step by step solution

01

Write Augmented Matrix

First, convert the given system of equations into an augmented matrix. The system is:\[-\frac{1}{3}x - \frac{1}{8}y + \frac{1}{6}z = -\frac{4}{3}\-\frac{2}{3}x - \frac{7}{8}y + \frac{1}{3}z = -\frac{23}{3} \-\frac{1}{3}x - \frac{5}{8}y + \frac{5}{6}z = 0\]The corresponding augmented matrix is:\[\begin{bmatrix}-\frac{1}{3} & -\frac{1}{8} & \frac{1}{6} & \vrule & -\frac{4}{3} \-\frac{2}{3} & -\frac{7}{8} & \frac{1}{3} & \vrule & -\frac{23}{3} \-\frac{1}{3} & -\frac{5}{8} & \frac{5}{6} & \vrule & 0 \end{bmatrix}\]
02

Perform Row Operations to Achieve Upper Triangular Form

Aim to create zeros below the first element of the first column. * Multiply the first row by 2 and add it to the second row to create a new second row:\[R_2 = R_2 + 2R_1\]* Multiply the first row by 1 and add it to the third row to create a new third row:\[R_3 = R_3 + R_1\]The operations result in the augmented matrix:\[\begin{bmatrix}-\frac{1}{3} & -\frac{1}{8} & \frac{1}{6} & \vrule & -\frac{4}{3} \0 & -\frac{11}{8} & \frac{1}{2} & \vrule & -14 \0 & -\frac{3}{8} & \frac{2}{3} & \vrule & -\frac{4}{3}\end{bmatrix}\]
03

Further Simplify the Matrix

Next, eliminate the element in the second column of the third row.* Multiply the second row by \(\frac{3}{11}\) and add to the third row: \[R_3 = R_3 + \frac{3}{11}R_2\]This results in:\[\begin{bmatrix}-\frac{1}{3} & -\frac{1}{8} & \frac{1}{6} & \vrule & -\frac{4}{3} \0 & -\frac{11}{8} & \frac{1}{2} & \vrule & -14 \0 & 0 & \frac{31}{44} & \vrule & -\frac{134}{11}\end{bmatrix}\]
04

Back Substitution

With the matrix in row-echelon form, perform back substitution:1. Solve for \(z\) using the third row: \[\frac{31}{44}z = -\frac{134}{11}\] Solving gives \(z = -\frac{44}{31}\times\frac{134}{11} = -\frac{4}{1}\)2. Substitute \(z\) into the second row to solve for \(y\): \[-\frac{11}{8}y + \frac{1}{2}(-4) = -14\] Solving gives \(y = 6\)3. Substitute \(y\) and \(z\) into the first row to solve for \(x\): \[-\frac{1}{3}x - \frac{1}{8}(6) + \frac{1}{6}(-4) = -\frac{4}{3}\] Solving gives \(x = 2\)
05

Verify the Solution

Substitute \(x = 2\), \(y = 6\), and \(z = -4\) back into the original equations to ensure they satisfy all three equations:\[-\frac{1}{3}(2) - \frac{1}{8}(6) + \frac{1}{6}(-4) = -\frac{4}{3},\-\frac{2}{3}(2) - \frac{7}{8}(6) + \frac{1}{3}(-4) = -\frac{23}{3},\-\frac{1}{3}(2) - \frac{5}{8}(6) + \frac{5}{6}(-4) = 0\]Each equation holds true, confirming the solution is correct.

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 dealing with systems of equations, an essential step is to convert the system into an augmented matrix. This matrix includes both the coefficients of the variables and the constants from the equations. For example, in the given system, we have three equations with variables \(x\), \(y\), and \(z\). Each equation contributes a row of numbers to the matrix. The coefficients of \(x\), \(y\), and \(z\) become the entries of the matrix, alongside the vertical bar which separates the coefficients from the constants on the right side of each equation.

This transformation is crucial because it prepares the equations for further manipulations using matrix techniques. The augmented matrix allows for a structured approach to solve the system by applying row operations and ultimately using methods like Gaussian elimination.
Row Operations
Row operations are a powerful tool used to simplify an augmented matrix into a form that is easier to solve. There are three types of row operations you can perform:
  • Swap two rows.
  • Multiply a row by a non-zero constant.
  • Add or subtract a multiple of one row to another row.

Using these operations, our goal is to obtain an upper triangular matrix, where all entries below the diagonal are zeros. This form makes the process of back substitution straightforward. For instance, in the solution provided, we performed operations like multiplying a row and adding it to another to eliminate the coefficients below the main diagonal. This systematic approach is what makes Gaussian elimination such an efficient method.
Back Substitution
After converting an augmented matrix to row-echelon form (upper triangular form), the next step is back substitution. This technique solves for the variables starting with the last row and working upward.

In our case, once the matrix is in an upper triangular form, we can easily find the value of \(z\) from the last row. With \(z\) known, we substitute it back into the previous rows to solve for \(y\) and then \(x\). Each step involves simple arithmetic to isolate the variable. It's like unraveling a mystery, step by step, until all variables are uncovered and the solution is fully revealed.
Systems of Equations
A system of equations, simply put, is a set of equations with multiple variables. The aim is to find values for these variables that satisfy all the equations simultaneously. Systems can be solved in various ways, including graphical, substitution, elimination methods, or matrix methods such as Gaussian elimination.

Understanding systems of equations is fundamental because they are widely applicable in various fields such as engineering, physics, and economics. Solving such systems efficiently can provide insights into complex problems where multiple factors interact. Through methods like Gaussian elimination, we not only find solutions but also gain deeper understanding of how mathematical interactions unfold.

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

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. A clothing store needs to order new inventory. It has three different types of hats for sale: straw hats, beanies, and cowboy hats. The straw hat is priced at \(\$ 13.99\) , the beanie at \(\$ 7.99,\) and the cowboy hat at \(\$ 14.49\) . If 100 hats were sold this past quarter, \(\$ 1,119\) was taken in by sales, and the amount of beanies sold was 10 more than cowboy hats, how many of each should the clothing store order to replace those already sold?

For the following exercises, use a calculator to solve the system of equations with matrix inverses. $$\begin{aligned}-\frac{1}{2} x-\frac{3}{2} y &=-\frac{43}{20} \\ \frac{5}{2} x+\frac{11}{5} y &=\frac{31}{4} \end{aligned}$$

For the following exercises, find the inverse of the given matrix. $$\left[\begin{array}{cccc}{1} & {0} & {1} & {0} \\ {0} & {1} & {0} & {1} \\\ {0} & {1} & {1} & {0} \\ {0} & {0} & {1} & {1}\end{array}\right]$$

For the following exercises, use this scenario: A health-conscious company decides to make a trail mix out of almonds, dried cranberries, and chocolate- covered cashews. The nutritional information for these items is shown in Table 1 . $$ \begin{array}{|c|c|c|c|} \hline & \text { Fat (g) } & \text { Protein (g) } & \text { Carbohydrates (g) } \\ \hline \text { Almonds (10) } & 6 & 2 & 3 \\ \hline \text { Cranberries (10) } & 0.02 & 0 & 8 \\ \hline \text { Cashews (10) } & 7 & 3.5 & 5.5 \\ \hline \end{array} $$ For the "energy-booster" mix, there are 1,000 pieces in the mix, containing \(145 \mathrm{~g}\) of protein and \(625 \mathrm{~g}\) of carbohydrates. If the number of almonds and cashews summed together is equivalent to the amount of cranberries, how many of each item is in the trail mix?

For the following exercises, solve the system by Gaussian elimination. $$ \begin{aligned} x+2 y-z &=1 \\ -x-2 y+2 z &=-2 \\ 3 x+6 y-3 z &=3 \end{aligned} $$

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.