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Magazine Chapter 4: Problem 40
Solve each system using the Gauss-Jordan elimination method. $$ \begin{array}{r} x+y=6 \\ -x+y=8 \end{array} $$
Short Answer
Expert verified
x = -1, y = 7
Step by step solution
01
Write the system as an augmented matrix
The given system of equations is: \[\begin{array}{r} x + y = 6 \ -x + y = 8 \end{array} \]Write it as an augmented matrix: \[\left[\begin{array}{cc|c}1 & 1 & 6 \-1 & 1 & 8 \end{array}\right]\]
02
Make the first element of the first column a 1
The first element in the first row and first column is already 1, so no row operation is needed. The matrix remains: \[\left[\begin{array}{cc|c}1 & 1 & 6 \-1 & 1 & 8 \end{array}\right]\]
03
Eliminate the first element of the second row
Add row 1 to row 2 to eliminate the -1 at position (2,1): \[R_2 \leftarrow R_2 + R_1\]\[\begin{array}{cc|c}1 & 1 & 6 \0 & 2 & 14 \end{array}\]
04
Make the leading coefficient of the second row a 1
Divide the entire second row by 2: \[R_2 \leftarrow \frac{1}{2}R_2\]\[\begin{array}{cc|c}1 & 1 & 6 \0 & 1 & 7 \end{array}\]
05
Eliminate the second element of the first row
Subtract row 2 from row 1 to eliminate the 1 at position (1,2): \[R_1 \leftarrow R_1 - R_2\]\[\begin{array}{cc|c}1 & 0 & -1 \0 & 1 & 7 \end{array}\]
06
Write the solution from the resulting matrix
The final matrix is in reduced row echelon form: \[\begin{array}{cc|c}1 & 0 & -1 \0 & 1 & 7 \end{array}\]This corresponds to the system: \[\begin{array}{r} x = -1 \ y = 7 \end{array}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solving Systems of Linear Equations
To solve systems of linear equations, we often seek a set of values for the variables that satisfy all equations simultaneously. This is a fundamental concept in linear algebra and involves expressing multiple linear equations, each representing a straight line in a multi-dimensional space. In our example, we had the system:
\[ \begin{array}{r} x + y = 6 \ -x + y = 8 \end{array} \] Here, we look for values of x and y that make both equations true at the same time.
\[ \begin{array}{r} x + y = 6 \ -x + y = 8 \end{array} \] Here, we look for values of x and y that make both equations true at the same time.
- There are various methods to tackle these problems: substitution, elimination, and the Gauss-Jordan elimination method.
- Gauss-Jordan elimination, the method we used in our example, systematically transforms the system's matrix into a simpler form.
Augmented Matrix
An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficient matrix with the constants on the right-hand side. For the system
\[ \begin{array}{r} x + y = 6 \ -x + y = 8 \end{array} \] the corresponding augmented matrix is:
\[ \begin{array}{cc|c} 1 & 1 & 6 \ -1 & 1 & 8 \end{array} \]
\[ \begin{array}{r} x + y = 6 \ -x + y = 8 \end{array} \] the corresponding augmented matrix is:
\[ \begin{array}{cc|c} 1 & 1 & 6 \ -1 & 1 & 8 \end{array} \]
- The vertical line separates the coefficients of variables from the constants.
- This form helps in applying row operations systematically to simplify the system.
Row Operations
Row operations are the fundamental tools for transforming the augmented matrix. They include:
- Swapping two rows.
- Multiplying a row by a nonzero scalar.
- Adding or subtracting a multiple of one row to another.
- We first wrote the augmented matrix and kept the first element of the first column at 1.
- We added the first row to the second row to eliminate the -1 at position (2,1), resulting in \[ \begin{array}{cc|c} 1 & 1 & 6 \ 0 & 2 & 14 \end{array} \]
- Next, we divided the second row by 2, making the leading coefficient of the second row a 1: \[ \begin{array}{cc|c} 1 & 1 & 6 \ 0 & 1 & 7 \end{array} \].
- Finally, we subtracted the second row from the first row to isolate x: \[ \begin{array}{cc|c} 1 & 0 & -1 \ 0 & 1 & 7 \end{array} \]
