Chapter 2: Problem 49
Solve the system of linear equations using the Gauss-Jordan elimination method. $$ \begin{array}{rr} x_{1}-x_{2}+3 x_{3}= & 14 \\ x_{1}+x_{2}+x_{3}= & 6 \\ -2 x_{1}-x_{2}+x_{3}= & -4 \end{array} $$
Short Answer
Expert verified
The solution for the given system of linear equations using the Gauss-Jordan elimination method is \(x_{1} = 4\), \(x_{2} = -1\), and \(x_{3} = 3\).
Step by step solution
01
Write down the augmented matrix
First, write down the given system of linear equations as an augmented matrix:
\[
\left[\begin{array}{rrr|r}
1 & -1 & 3 & 14 \\
1 & 1 & 1 & 6 \\
-2 & -1 & 1 & -4
\end{array}\right]
\]
02
Swap rows if necessary
Next, we need to make sure that the first element of the first row is non-zero. Since it is already non-zero, we don't need to swap any rows.
03
Make the first element of the first row equal to 1
The first element of the first row (1) is already equal to 1, so this step is already done.
04
Make the other elements in the first column equal to 0
We want the first column to look like this:
\[
\left[\begin{array}{r}
1 \\
0 \\
0
\end{array}\right]
\]
To do that, we will perform the following row operations:
1. Subtract the first row from the second row
2. Add twice the first row to the third row
These operations result in the following matrix:
\[
\left[\begin{array}{rrr|r}
1 & -1 & 3 & 14 \\
0 & 2 & -2 & -8 \\
0 & -3 & 7 & 24
\end{array}\right]
\]
05
Repeat steps 2-4 for the second row
The first element of the second row is non-zero, so we can proceed by making the second element of the second row equal to 1. Divide the second row by 2 to achieve this:
\[
\left[\begin{array}{rrr|r}
1 & -1 & 3 & 14 \\
0 & 1 & -1 & -4 \\
0 & -3 & 7 & 24
\end{array}\right]
\]
Now, make other elements in the second column equal to 0 by performing the following row operations:
1. Add the second row to the first row
2. Add three times the second row to the third row
\[
\left[\begin{array}{rrr|r}
1 & 0 & 2 & 10 \\
0 & 1 & -1 & -4 \\
0 & 0 & 4 & 12
\end{array}\right]
\]
06
Repeat step 2-4 for the third row
The first element of the third row is non-zero, so we can proceed by making the third element of the third row equal to 1. Divide the third row by 4 to achieve this:
\[
\left[\begin{array}{rrr|r}
1 & 0 & 2 & 10 \\
0 & 1 & -1 & -4 \\
0 & 0 & 1 & 3
\end{array}\right]
\]
Now, make other elements in the third column equal to 0 by performing the following row operations:
1. Subtract two times the third row from the first row
2. Add the third row to the second row
\[
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 4 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3
\end{array}\right]
\]
The matrix is now in row-reduced echelon form, and we can use it to find the values of the variables:
\(x_{1} = 4\)
\(x_{2} = -1\)
\(x_{3} = 3\)
So, the solution for the given system of linear equations is \(x_{1} = 4\), \(x_{2} = -1\), and \(x_{3} = 3\).
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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 a collection of two or more linear equations involving the same set of variables. Each equation represents a line, plane, or hyperplane, and the solution to the system is a point or set of points where these objects intersect. For example, in a two-dimensional system with variables \(x\) and \(y\), each equation might take the form \(ax + by = c\). The challenge is to find values for these variables that satisfy all the equations simultaneously.
To solve such a system, one can use various methods, including substitution, elimination, and matrix-based approaches like Gauss-Jordan elimination. Understanding and visualizing the intersection of lines or planes can help grasp how solutions might appear as a single point, line, or infinite set of solutions depending on the conditions given by the equations.
To solve such a system, one can use various methods, including substitution, elimination, and matrix-based approaches like Gauss-Jordan elimination. Understanding and visualizing the intersection of lines or planes can help grasp how solutions might appear as a single point, line, or infinite set of solutions depending on the conditions given by the equations.
Augmented Matrix
An augmented matrix is a crucial tool in solving systems of linear equations, especially when using matrix methods like Gauss-Jordan elimination. It combines the coefficients of the variables and the constants from the equations into a single matrix.
Here's how you form an augmented matrix:
Here's how you form an augmented matrix:
- List the coefficients of each variable in a row, aligning coefficients of each equation?
- Add a vertical line or a different section to the last column to include the constant terms from the right-hand side of each equation.
Row Operations
Row operations are essential procedures within the Gauss-Jordan elimination method. These operations help transform an augmented matrix into a simpler form, ideally the row-reduced echelon form. There are three basic row operations:
- Swapping Rows: You can interchange two rows as necessary to simplify calculations or proceed with eliminating variables.
- Scalar Multiplication: A row can be multiplied by a non-zero constant. This operation is often used to turn row elements into 1s for easier eliminations.
- Row Addition: Add or subtract a multiple of one row to another. This helps in creating zeros below or above the leading 1s in a column.
Row-Reduced Echelon Form
The row-reduced echelon form (RREF) of a matrix is one of the final steps in solving a system of linear equations using the Gauss-Jordan method. The RREF is a specific matrix format where each leading entry (the first non-zero number from the left) in a row is 1, and above or below each leading 1, all other elements in that pivot column are zero. Additionally,
- The leading 1 in a given row is to the right of the leading 1 in the row above.
- Any row containing all zeroes is at the bottom of the matrix.