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Chapter 2: Problem 23

Solve the system of linear equations, using the Gauss-Jordan elimination method. $$ \begin{aligned} 2 x_{1}-x_{2}+x_{3} &=-4 \\ 3 x_{1}-\frac{3}{2} x_{2}+\frac{3}{2} x_{3} &=-6 \\ -6 x_{1}+3 x_{2}-3 x_{3} &=12 \end{aligned} $$

Short Answer

Expert verified
The solution to the given system of linear equations using the Gauss-Jordan elimination method is \(x_{1} = -1, x_{2} = 1,\) and \(x_{3} = 1\).

Step by step solution

01

Write the augmented matrix

Write the augmented matrix for the given system of linear equations: $$ \left[\begin{array}{ccc|c} 2 & -1 & 1 & -4 \\ 3 & -\frac{3}{2} & \frac{3}{2} & -6 \\ -6 & 3 & -3 & 12 \end{array}\right] $$
02

Perform Gauss-Jordan elimination

> Begin by manipulating the matrix to achieve its "reduced row-echelon form": 1. Divide row 1 by 2: $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{1}{2} & -2 \\ 3 & -\frac{3}{2} & \frac{3}{2} & -6 \\ -6 & 3 & -3 & 12 \end{array}\right] $$ 2. Add -3 times row 1 to row 2 and 6 times row 1 to row 3: $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{1}{2} & -2 \\ 0 & -\frac{3}{2} & \frac{3}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ 3. Multiply row 2 by -\(\frac{2}{3}\): $$ \left[\begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{1}{2} & -2 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ 4. Add \(\frac{1}{2}\) times row 2 to row 1: $$ \left[\begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ Now, we have obtained the reduced row-echelon form of the matrix.
03

Obtain the solutions

From the reduced row-echelon form, we can read off the solutions: $$ \begin{aligned} x_{1} &= -1 \\ x_{2} &= 1 \\ x_{3} &= 1 \end{aligned} $$ Thus, the solution to the given system of linear equations is \(x_{1} = -1, x_{2} = 1,\) and \(x_{3} = 1\).

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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 like a puzzle where each equation is a piece. These equations involve multiple variables, such as \(x_1, x_2,\) and \(x_3\) in our example. The goal is to find the values for these variables that satisfy all the equations simultaneously.
These systems can have:
  • Exactly one solution, where all equations intersect at a single point.
  • No solution, if there is no point that satisfies all the equations together.
  • Infinitely many solutions, like parallel lines that overlap completely.
Using methods like substitution might help in simple cases, but for more complex puzzles, techniques such as Gauss-Jordan elimination are your best bet.
Augmented Matrix
An augmented matrix is a neat way to organize a system of linear equations. Imagine taking all your equations and turning them into a table format, showing only the numbers and leaving the variables behind. It looks like this for the given system:\[\begin{array}{ccc|c}2 & -1 & 1 & -4 \3 & -\frac{3}{2} & \frac{3}{2} & -6 \-6 & 3 & -3 & 12\end{array}\]The lines of numbers separated by pipes represent the coefficients from each equation, plus a column for the equalities.
The augmented matrix is important because it helps simplify calculations and keeps everything organized while applying methods like Gauss-Jordan elimination.
Reduced Row-Echelon Form
Achieving the reduced row-echelon form (RREF) is like organizing a pile of scrambled puzzles into a clear, solved picture. Using row operations, which are simple swaps and transformations, the matrix moves from a jumble of numbers to a precise format where solutions are easily seen:\[\begin{array}{ccc|c}1 & 0 & 0 & -1 \0 & 1 & -1 & 0 \0 & 0 & 0 & 0\end{array}\]In RREF, the leading entry in each non-zero row is 1, and it's arranged neatly in diagonal form, like stepping stones across the rows. This structure allows for easy reading of solutions directly from the matrix.
This form is powerful because it reveals the system's nature鈥攚hether it has a unique solution, infinitely many, or none.
Solution of Equations
Finding the solution to a system of linear equations is the ultimate goal of processes like Gauss-Jordan elimination. Once the matrix is in reduced row-echelon form, the solutions are clearly seen by simply observing the matrix.
In the example given, the solutions can be immediately written as:
  • \(x_1 = -1\)
  • \(x_2 = 1\)
  • \(x_3 = 1\)
This creates a clear picture of which values for \(x_1, x_2,\) and \(x_3\) satisfy all the original equations.
Sometimes, the solution might indicate infinite options or none at all, depending on whether rows in the RREF matrix result in contradictions or identities.

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Most popular questions from this chapter

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. If \(A\) and \(B\) are matrices of the same size and \(c\) is a scalar, then \(c(A+B)=c A+c B\).

The Campus Bookstore's inventory of books is Hardcover: textbooks, 5280 ; fiction, 1680 ; nonfiction, \(2320 ;\) reference, 1890 Paperback: fiction, 2810; nonfiction, 1490; reference, \(2070 ;\) textbooks, 1940 The College Bookstore's inventory of books is Hardcover: textbooks, 6340; fiction, 2220; nonfiction, 1790 ; reference, 1980 Paperback: fiction, 3100; nonfiction, 1720 ; reference, \(2710 ;\) textbooks, 2050 a. Represent Campus's inventory as a matrix \(A\). b. Represent College's inventory as a matrix \(B\). c. The two companies decide to merge, so now write a matrix \(C\) that represents the total inventory of the newly amaluomated \(\mathrm{com} \mathrm{n}\)

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.

Matrix \(A\) is an input-output matrix associated with an economy, and matrix \(D\) (units in millions of dollars) is a demand vector. In each problem,find the final outputs of each industry such that the demands of industry and the consumer sector are met. $$ A=\left[\begin{array}{lll} 0.2 & 0.4 & 0.1 \\ 0.3 & 0.2 & 0.1 \\ 0.1 & 0.2 & 0.2 \end{array}\right] \text { and } D=\left[\begin{array}{r} 6 \\ 8 \\ 10 \end{array}\right] $$

Let $$ A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] $$ a. Find \(A^{-1}\) if it exists. b. Find a necessary condition for \(A\) to be nonsingular. c. Verify that \(A A^{-1}=A^{-1} A=I\).
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