/*! 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 30 Solve the system of linear equat... [FREE SOLUTION] | 91影视

91影视

Solve the system of linear equations, using the Gauss-Jordan elimination method. $$ \begin{array}{rr} 3 x-9 y+6 z= & -12 \\ x-3 y+2 z= & -4 \\ 2 x-6 y+4 z= & 8 \end{array} $$

Short Answer

Expert verified
The given system of linear equations has no solution, as the Reduced Row Echelon Form (RREF) yields an inconsistent equation: \(0x + 0y + 0z = 16\).

Step by step solution

01

1. Form the Augmented Matrix

First, let's represent the given system of linear equations as an augmented matrix: $$ \left[\begin{array}{rrr|r} 3 & -9 & 6 & -12 \\ 1 & -3 & 2 & -4 \\ 2 & -6 & 4 & 8 \end{array}\right] $$
02

2. Apply Row Operations to Reach RREF

We will apply row operations to simplify the augmented matrix. First, we want a '1' in the top left corner of our matrix: Row1 = Row1 / 3: $$ \left[\begin{array}{rrr|r} 1 & -3 & 2 & -4 \\ 1 & -3 & 2 & -4 \\ 2 & -6 & 4 & 8 \end{array}\right] $$ Subtract Row1 from Row2: $$ \left[\begin{array}{rrr|r} 1 & -3 & 2 & -4 \\ 0 & 0 & 0 & 0 \\ 2 & -6 & 4 & 8 \end{array}\right] $$ Subtract 2 times Row1 from Row3: $$ \left[\begin{array}{rrr|r} 1 & -3 & 2 & -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 16 \\ \end{array}\right] $$
03

3. Solve for x, y, and z from the RREF

From the resulting augmented matrix, we can see that the last row corresponds to the equation 鈥0x + 0y + 0z = 16鈥. This equation has no solution since the left side will always equal 0 and the right side is equal to 16. Since this equation is inconsistent, we can conclude that the given system of linear 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 91影视!

Key Concepts

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

System of Linear Equations
A system of linear equations is a collection of one or more linear equations involving the same set of variables. In this particular exercise, we are working with a set of three linear equations that involve the variables \(x\), \(y\), and \(z\). These equations are:
  • \(3x - 9y + 6z = -12\)
  • \(x - 3y + 2z = -4\)
  • \(2x - 6y + 4z = 8\)
The goal with systems like these is to find values for each of the variables that satisfy all the equations simultaneously. There are several methods to solve such systems, and in this case, we're utilizing the Gauss-Jordan elimination method. This method helps simplify these equations into a form that makes the solutions more apparent.
Augmented Matrix
To solve a system of linear equations more systematically, we often convert it into an augmented matrix. This matrix combines all coefficients of the variables, with an additional column that represents the constants from the right-hand side of the equations. For our system, the augmented matrix looks like this:\[\begin{array}{ccc|c}3 & -9 & 6 & -12 \1 & -3 & 2 & -4 \2 & -6 & 4 & 8 \\end{array}\]The first three columns correspond to the coefficients of \(x\), \(y\), and \(z\), and the last column contains the constants from the equations. Using this matrix form facilitates the application of row operations, which are crucial for simplifying the system into a more easily solvable form.
Row Operations
Row operations are techniques used in linear algebra to manipulate rows of a matrix. These operations include swapping rows, multiplying a row by a nonzero scalar, and adding or subtracting the multiples of one row from another. In the Gauss-Jordan elimination process, row operations are employed to simplify the augmented matrix into reduced row-echelon form (RREF). This process involves steps like:
  • Getting a '1' in the pivot position 鈥 typically, the top-left position of the matrix.
  • Making zeros below and above this pivot for simpler variables isolation.
Through steps such as dividing Row 1 by 3, subtracting Row 1 from Row 2, and subtracting 2 times Row 1 from Row 3, we systematically transform the matrix to assess the presence of solutions.
Inconsistent System
An inconsistent system of equations is one where there are no solutions. This typically occurs when the matrix operations yield a row in the form \([0 \, 0 \, 0 | c]\) where \(c\) is a non-zero constant. This form implies that the equations present a contradiction, such as stating "0 equals a non-zero number."In our exercise, the last row of the matrix simplifies to \([0 \, 0 \, 0 | 16]\). This indicates that no combination of \(x\), \(y\), and \(z\) can satisfy all the equations simultaneously, confirming that the system is inconsistent. In real-world applications, this kind of result might indicate parallel lines that never intersect or other scenarios without a logical solution.

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

The total output of loudspeaker systems of the Acrosonic Company at their three production facilities for May and June is given by the matrices \(A\) and \(B\), respectively, where $$ \begin{array}{l} \left.\begin{array}{ccccc} & \text { Model } & \text { Model } & \text { Model } & \text { Model } \\ & \text { A } & \text { B } & \text { C } & \text { D } \\ \text { Location I } & {[320} & 280 & 460 & 280 \\ A= & \text { Location II } & 480 & 360 & 580 & 0 \\ \text { Location III } & 540 & 420 & 200 & 880 \end{array}\right]\\\ \left.\begin{array}{ccccc} & \text { Model } & \text { Model } & \text { Model } & \text { Model } \\ & \text { A } & \text { B } & \text { C } & \text { D } \\ \text { Location I } & 210 & 180 & 330 & 180 \\ B=\text { Location II } & 400 & 300 & 450 & 40 \\ \text { Location III } & 420 & 280 & 180 & 740 \end{array}\right] \end{array} $$ The unit production costs and selling prices for these loudspeakers are given by matrices \(C\) and \(D\), respectively, where \(\begin{array}{c}\text { Model A } \\ C=\begin{array}{l}\text { Model B } \\\ \text { Model C } \\ \text { Model D }\end{array} & {\left[\begin{array}{l}120 \\ 180 \\ 260 \\ 500\end{array}\right] \text { and } \quad D=\begin{array}{c}\text { Model A } \\ \text { Model B } \\\ \text { Model C } \\ \text { Model D }\end{array}} & {\left[\begin{array}{l}160 \\ 250 \\ 350 \\\ 700\end{array}\right]}\end{array}\) Compute the following matrices and explain the meaning of the entries in each matrix. a. \(A C\) b. \(A D\) c. \(B C\) d. \(B D\) e. \((A+B) C\) f. \((A+B) D\) g. \(A(D-C)\) h. \(B(D-C)\) i. \((A+B)(D-C)\)

\(e\) $$ A=\left[\begin{array}{rr} 2 & 2 \\ -2 & -2 \end{array}\right] $$ Show that \(A^{2}=0\). Compare this with the equation \(a^{2}=0\), where \(a\) is a real number.

Let $$ A=\left[\begin{array}{rr} 2 & 3 \\ -4 & -5 \end{array}\right] $$ a. Find \(A^{-1}\). b. Show that \(\left(A^{-1}\right)^{-1}=A\).

The problems in exercise correspond to those in exercises 15-27, Section 2.1. Use the results of your previous work to help you solve these problems. The annual returns on Sid Carrington's three investments amounted to $$\$ 21,600$$: \(6 \%\) on a savings account, \(8 \%\) on mutual funds, and \(12 \%\) on bonds. The amount of Sid's investment in bonds was twice the amount of his investment in the savings account, and the interest earned from his investment in bonds was equal to the dividends he received from his investment in mutual funds. Find how much money he placed in each type of investment.

Mr. and Mrs. Garcia have a total of $$\$ 100,000$$ to be invested in stocks, bonds, and a money market account. The stocks have a rate of return of \(12 \% /\) year, while the bonds and the money market account pay \(8 \% /\) year and \(4 \%\) year, respectively. The Garcias have stipulated that the amount invested in the money market account should be equal to the sum of \(20 \%\) of the amount invested in stocks and \(10 \%\) of the amount invested in bonds. How should the Garcias allocate their resources if they require an annual income of $$\$$ 10,000$ from their investments?

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.