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Chapter 2: Problem 44

Solve the system of linear equations using the Gauss-Jordan elimination method. $$ \begin{array}{rr} 2 x+4 y-6 z= & 38 \\ x+2 y+3 z= & 7 \\ 3 x-4 y+4 z= & -19 \end{array} $$

Short Answer

Expert verified
The solution to the given system of linear equations using the Gauss-Jordan elimination method is \(x = -3\), \(y = 8\), and \(z = -2\).

Step by step solution

01

Write the equation in matrix form

First, we rewrite the given system of linear equations as a matrix (called the coefficient matrix). We obtain: \[ \begin{bmatrix} 2 & 4 & -6 \\ 1 & 2 & 3 \\ 3 & -4 & 4 \end{bmatrix} \]
02

Convert matrix to an augmented matrix

Next, we create an augmented matrix by adding a column for the constants on the right-hand side of the given system of equations: \[ \begin{bmatrix} 2 & 4 & -6 & | & 38 \\ 1 & 2 & 3 & | & 7 \\ 3 & -4 & 4 & | & -19 \end{bmatrix} \]
03

Perform Gauss-Jordan elimination to reach RREF

Now we perform a series of row operations on the augmented matrix in order to obtain its reduced row echelon form (RREF). 1. Divide Row 1 by `2`: \[ \begin{bmatrix} 1 & 2 & -3 & | & 19 \\ 1 & 2 & 3 & | & 7 \\ 3 & -4 & 4 & | & -19 \end{bmatrix} \] 2. Replace Row 2 with `Row 2 - Row 1` and replace Row 3 with `Row 3 - 3 脳 Row 1`: \[ \begin{bmatrix} 1 & 2 & -3 & | & 19 \\ 0 & 0 & 6 & | & -12 \\ 0 & -10 & 13 & | & -76 \end{bmatrix} \] 3. Divide Row 2 by `6` and Row 3 by `-10`: \[ \begin{bmatrix} 1 & 2 & -3 & | & 19 \\ 0 & 0 & 1 & | & -2 \\ 0 & 1 & -\frac{13}{10} & | & 76/10 \end{bmatrix} \] 4. Replace Row 1 with `Row 1 + 3 脳 Row 2` and Row 3 with `Row 3 - 2 脳 Row 2`: \[ \begin{bmatrix} 1 & 2 & 0 & | & 13 \\ 0 & 0 & 1 & | & -2 \\ 0 & 1 & 0 & | & 8 \end{bmatrix} \] 5. Replace Row 1 with `Row 1 - 2 脳 Row 3`: \[ \begin{bmatrix} 1 & 0 & 0 & | & -3 \\ 0 & 0 & 1 & | & -2 \\ 0 & 1 & 0 & | & 8 \end{bmatrix} \]
04

Interpret the reduced matrix

In reduced form, our matrix represents the following system of linear equations: \[ \begin{cases} x= -3 \\ y= 8 \\ z= -2 \end{cases} \]
05

Write the solution

The solution of the system of linear equations is: \[ x = -3, y = 8, z = -2 \]

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Key Concepts

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

Systems of Linear Equations
When faced with a system of linear equations, we're essentially looking at multiple linear equations that are all interconnected. Finding solutions to these systems involves determining the values for the variables that make all equations true simultaneously.

For example, a system may have two or more equations involving the same variables, such as the set of equations provided in the exercise. We aim to find the values for the variables x, y, and z that will satisfy all three equations at once. The beauty of linear systems is that they can be visualized geometrically as lines or planes in space, and their intersection points correspond to the solutions. The Gauss-Jordan elimination method, one of several techniques available, transforms these systems into a much more manageable form called the reduced row echelon form, from which the solutions can be directly read.
Reduced Row Echelon Form
With Gauss-Jordan elimination, our goal is to simplify the system to its reduced row echelon form (RREF). This tidy state is characterized by each leading entry of a row being 1 (if it's not all zeroes), and all elements above and below the leading 1 being zeros. Furthermore, each leading 1 must be to the right of any leading 1s in the rows above.

The properties of RREF make it a powerful tool in linear algebra. When a matrix is in this form, the solution of the system of equations it represents becomes clear: each row corresponds to an equation with a leading variable that can be solved directly. If a row is filled with zeros except for the last element, it means the system has no solution. If there's a row of complete zeros, it indicates the possibility of infinite solutions depending on other rows.
Matrix Operations
Matrices aren't just arrays of numbers; they're capable of undergoing various operations that help us solve systems of equations. In the context of the Gauss-Jordan elimination method, we primarily use three elementary row operations:
  • Row Switching - swapping the positions of two rows.
  • Row Multiplication - multiplying an entire row by a non-zero constant.
  • Row Addition - adding or subtracting the multiples of one row to another.

These operations preserve the solutions of the system but make the matrix more manageable. We apply these strategically in the steps of our method to gradually coerce the matrix into RREF. It's essential to perform these steps carefully, as improper application can lead to incorrect solutions.
Augmented Matrix
An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficients of the variables and the constants from the right-hand side of the equations into a single matrix. This is done by placing a vertical bar (or partition) to separate the coefficient's part of the matrix from the constants.

The augmented matrix simplifies operations by containing all the necessary information to perform row reductions. As we manipulate the augmented matrix through Gauss-Jordan elimination, we simultaneously change the coefficients and constants, keeping the integrity of the equations intact. This allows us to march step by step towards an easier-to-read form that ultimately reveals the system's solution.

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Most popular questions from this chapter

Joan and Dick spent 2 wk (14 nights) touring four cities on the East Coast- Boston, New York, Philadelphia, and Washington. They paid $$\$ 120$$, $$\$ 200$$, $$\$ 80$$, and $$\$ 100$$ per night for lodging in each city, respectively, and their total hotel bill came to $$\$ 2020$$. The number of days they spent in New York was the same as the total number of days they spent in Boston and Washington, and the couple spent 3 times as many days in New York as they did in Philadelphia. How many days did Joan and Dick stay in each city?

William's and Michael's stock holdings are given by the matrix C GI (B) 1 \(A=\begin{array}{l}\text { William } \\ \text { Michael }\end{array}\left[\begin{array}{lllr}200 & 300 & 100 & 200 \\ 100 & 200 & 400 & 0\end{array}\right]\) At the close of trading on a certain day, the prices (in dollars per share) of the stocks are given by the matrix $$ \begin{array}{c} \mathrm{BAC} \\ B=\mathrm{GM} & 54 \\ \mathrm{IBM} & 48 \\ \mathrm{TRW} & 98 \\ 82 \end{array} $$ a. Find \(A B\) b. Explain the meaning of the entries in the matrix \(A B\).

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. $$ \begin{aligned} x+2 y+z &=b_{1} \\ x+y+z &=b_{2} \\ 3 x+y+z &=b_{3} \\ \text { where } & \text { (i) } b_{1}=7, b_{2}=4, b_{3}=2 \\ \text { and } & \text { (ii) } b_{1}=5, b_{2}=-3, b_{3}=-1 \end{aligned} $$

The amount of money raised by charity I, charity II, and charity III (in millions of dollars) in each of the years 2006,2007, and 2008 is represented by the matrix \(A\) : $$ A=\begin{array}{cccc} & \mathrm{I} & \mathrm{II} & \mathrm{III} \\ 2006 & {\left[\begin{array}{ccc} 18.2 & 28.2 & 40.5 \\ 19.6 & 28.6 & 42.6 \\ 20.8 & 30.4 & 46.4 \end{array}\right]} \\ 2007 & 2008 \end{array} $$ On average, charity I puts \(78 \%\) toward program cost, charity II puts \(88 \%\) toward program cost, and charity III puts \(80 \%\) toward program cost. Write a \(3 \times 1\) matrix \(B\) reflecting the percentage put toward program cost by the charities. Then use matrix multiplication to find the total amount of money put toward program cost in each of the 3 yr by the charities under consideration.

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. A system of linear equations having fewer equations than variables has no solution, a unique solution, or infinitely many solutions.
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