Chapter 8: Problem 6
In Problems 1-20, use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. $$ \begin{array}{r} x_{1}+2 x_{2}-x_{3}=0 \\ 2 x_{1}+x_{2}+2 x_{3}=9 \\ x_{1}-x_{2}+x_{3}=3 \end{array} $$
Short Answer
Expert verified
The solution is \(x_1 = 6\), \(x_2 = -3\), and \(x_3 = 0\).
Step by step solution
01
Write the Augmented Matrix
First, convert the system of equations into an augmented matrix. For the given system: \(\begin{bmatrix} 1 & 2 & -1 & | & 0 \ 2 & 1 & 2 & | & 9 \ 1 & -1 & 1 & | & 3 \end{bmatrix}\).
02
Make the Leading Entry of First Row to 1
The leading entry of the first row is already 1. No changes are needed for Row 1.
03
Eliminate First Column Entries Below the Leading 1
Use Row 1 to eliminate the below entries by performing Row 2 - 2*Row 1 and Row 3 - Row 1:- Row 2 becomes \([0, -3, 4, 9] \).- Row 3 becomes \([0, -3, 2, 3] \).
04
Make the Leading Entry in the Second Row to 1
Currently, the leading entry in Row 2 is -3. Divide the entire row by -3 to turn the leading entry to 1: Row 2 becomes \([0, 1, -\frac{4}{3}, -3] \).
05
Eliminate the Second Column Entries Above and Below the Leading Entry
Use Row 2 to eliminate the other entries in the second column:- Row 1 becomes [1, 0, \(\frac{1}{3}\), 6].- Row 3 becomes [0, 0, -\(\frac{2}{3}\), 0].
06
Make the Leading Entry in the Third Row to 1
The leading entry in Row 3 is -\(\frac{2}{3}\). Divide the entire row by -\(\frac{2}{3}\) to get 1: Row 3 becomes \([0, 0, 1, 0] \).
07
Eliminate the Third Column Entries Above the Leading Entry
Using Row 3, eliminate the third column entries above the leading 1:- Correct Row 1 to \([1, 0, 0, 6] \).- Row 2 remains \([0, 1, 0, -3] \).
08
Write the Solution from the Row-Echelon Form
Read the solutions directly from the matrix:- \(x_1 = 6\), \(x_2 = -3\), \(x_3 = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gauss-Jordan elimination
Gauss-Jordan elimination is a method used in linear algebra to solve systems of linear equations. It is an extension of Gaussian elimination. This technique simplifies matrices to row-reduced echelon form. This specific form makes finding solutions straightforward, as each row corresponds to a simpler equation.
During Gauss-Jordan elimination, we perform row operations on the augmented matrix until we achieve row-echelon form and then continue to modify it to reduce further to just having zeros below and above the pivot positions. The key steps involve:
During Gauss-Jordan elimination, we perform row operations on the augmented matrix until we achieve row-echelon form and then continue to modify it to reduce further to just having zeros below and above the pivot positions. The key steps involve:
- Making leading entries (also known as pivot elements) of each row equal to 1.
- Eliminating all other entries in the pivot columns by adding or subtracting suitable multiples of the pivot row.
augmented matrix
The concept of an augmented matrix offers a practical way to handle systems of linear equations. By creating an augmented matrix, we combine the coefficients of the variables and the constants of the equations into a single matrix.
For a system of equations:
It provides a compact form facilitating easier manipulation during elimination procedures, such as Gaussian or Gauss-Jordan elimination. Using an augmented matrix streamlines the transition from manipulating algebraic equations directly to performing matrix operations, which significantly speeds up the process of finding solutions.
For a system of equations:
- Each row in the augmented matrix corresponds to one equation from the system.
- Columns represent the coefficients of each variable and a separate column for the constants on the right-hand side.
It provides a compact form facilitating easier manipulation during elimination procedures, such as Gaussian or Gauss-Jordan elimination. Using an augmented matrix streamlines the transition from manipulating algebraic equations directly to performing matrix operations, which significantly speeds up the process of finding solutions.
row-echelon form
Row-echelon form is the result of applying Gaussian elimination to a matrix. In this form, the structure of the matrix will greatly simplify. Each leading entry of a row (the first non-zero number from the left, also called a pivot) must be to the right of the leading entry of any row above it. If a row is all zeros, it should be at the bottom of the matrix.
Characteristics of this form include:
This form is a stepping stone to further simplifications, particularly when moving towards reduced row-echelon form in Gauss-Jordan elimination. Achieving this form is crucial, as it simplifies back-substitution when solving for variables, leading ultimately to the direct extraction of solutions from the matrix.
Characteristics of this form include:
- The first non-zero value in a row, called a pivot, is always 1, allowing easy eliminations in its column.
- All entries below each pivot are zeros.
This form is a stepping stone to further simplifications, particularly when moving towards reduced row-echelon form in Gauss-Jordan elimination. Achieving this form is crucial, as it simplifies back-substitution when solving for variables, leading ultimately to the direct extraction of solutions from the matrix.