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Magazine Chapter 18: Problem 9
The questions are all straightforward and based on the work covered. You will have no trouble. For the following set of simultaneous equations: (a) form the augmented coefficient matrix (b) solve the equations by Gaussian elimination. $$ \begin{gathered} x_{1}+2 x_{2}+3 x_{3}=5 \\ 3 x_{1}-x_{2}+2 x_{3}=8 \\ 4 x_{1}-6 x_{2}-4 x_{3}=-2 \end{gathered} $$
Short Answer
Expert verified
The solutions are \(x_1 = -1\), \(x_2 = -3\), and \(x_3 = 4\).
Step by step solution
01
Write Augmented Coefficient Matrix
First, we need to create the augmented coefficient matrix for the given system of linear equations. We shall represent each equation as a row:1. The equation \( x_1 + 2x_2 + 3x_3 = 5 \) can be represented as the row: \([1, 2, 3 | 5] \).2. The equation \( 3x_1 - x_2 + 2x_3 = 8 \) as the row: \([3, -1, 2 | 8] \).3. The equation \( 4x_1 - 6x_2 - 4x_3 = -2 \) as: \([4, -6, -4 | -2] \).Thus, the augmented matrix is: \[\begin{bmatrix}1 & 2 & 3 & | & 5 \3 & -1 & 2 & | & 8 \4 & -6 & -4 & | & -2\end{bmatrix}\]
02
Start Gaussian Elimination
The goal of Gaussian elimination is to transform the matrix into an upper triangular form. We start with the first pivot element in the first row.Current matrix:\[\begin{bmatrix}1 & 2 & 3 & | & 5 \3 & -1 & 2 & | & 8 \4 & -6 & -4 & | & -2\end{bmatrix}\]We first eliminate the first column below the pivot in the first row. This is done by replacing row 2 with \(\text{row 2} - 3 \times \text{row 1}\) and row 3 with \(\text{row 3} - 4 \times \text{row 1}\).
03
Perform Row Operations
1. For row 2: - Apply: \([3, -1, 2, 8] - 3 \times [1, 2, 3, 5]\) - Result: \([0, -7, -7, -7]\).2. For row 3: - Apply: \([4, -6, -4, -2] - 4 \times [1, 2, 3, 5]\) - Result: \([0, -14, -16, -22]\).The matrix becomes:\[\begin{bmatrix}1 & 2 & 3 & | & 5 \0 & -7 & -7 & | & -7 \0 & -14 & -16 & | & -22\end{bmatrix}\]
04
Continue Upper Triangular Transformation
Next, we need to eliminate the element in the second row, third column by making the row below zero using the row above it. We will focus on column 2 below the diagonal.We make the element in row 3, column 2 zero by replacing row 3 with \(\text{row 3} - 2 \times \text{row 2}\).
05
Apply Row Operation on Row 3
1. For row 3: - Apply: \([0, -14, -16, -22] - 2 \times [0, -7, -7, -7]\) - Result: \([0, 0, -2, -8]\).The updated matrix is:\[\begin{bmatrix}1 & 2 & 3 & | & 5 \0 & -7 & -7 & | & -7 \0 & 0 & -2 & | & -8\end{bmatrix}\]
06
Back Substitution
Now we have an upper triangular matrix. We perform back substitution to find the solutions.From the third row: \(-2x_3 = -8\) implies \(x_3 = 4\).Substituting \(x_3 = 4\) in the second row: \(-7x_2 - 7(4) = -7\), which simplifies to \(-7x_2 = 21\), giving \(x_2 = -3\).Finally, substituting \(x_2 = -3\) and \(x_3 = 4\) in the first row:\(x_1 + 2(-3) + 3(4) = 5\) simplifies to \(x_1 = -1\).
07
Complete Solution
The solutions are:- \(x_1 = -1\)- \(x_2 = -3\)- \(x_3 = 4\)Hence, the simultaneous equations are solved, providing the values for each variable.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Augmented Matrix
When solving systems of simultaneous equations, creating an augmented matrix is an essential first step. An augmented matrix is a compact way of representing a system of equations. It combines the coefficients of the variables and the constants from each equation into a single matrix.
Here's how you construct an augmented matrix:
Here's how you construct an augmented matrix:
- Each row in the matrix corresponds to an equation in the system.
- The columns are arranged from left to right, representing the coefficients of each variable followed by the constant term, separated by a vertical line.
- \( x_1 + 2x_2 + 3x_3 = 5 \)
- \( 3x_1 - x_2 + 2x_3 = 8 \)
- \( 4x_1 - 6x_2 - 4x_3 = -2 \)
Upper Triangular Form
To solve a system of equations using Gaussian elimination, one key step is converting the augmented matrix into an upper triangular form. This involves simplifying the matrix until all elements below the main diagonal are zero.
The process includes:
The process includes:
- Using row operations to create zeros beneath each pivot position. These operations include row swapping, scaling a row by a scalar, and adding or subtracting rows from one another.
- Focusing first on the leftmost column, and then proceeding to the right for subsequent columns.
- Each pivot step should transform the matrix closer to having zeroes below the pivot in its column.
Back Substitution
Once the augmented matrix is in upper triangular form, the next step in Gaussian elimination is back substitution. This technique is used to find the values of the unknowns in the system of equations, starting from the last row of the matrix.
Here's how back substitution works:
The solution to the system would be \[x_1 = -1, \, x_2 = -3, \, x_3 = 4.\]
Here's how back substitution works:
- Begin with the last row, which generally will contain only one variable. Solve for that variable directly.
- Substitute the found values back into the preceding rows, one at a time, to find the other unknowns.
- Continue this process until all variables are determined.
The solution to the system would be \[x_1 = -1, \, x_2 = -3, \, x_3 = 4.\]
Simultaneous Equations
Simultaneous equations are sets of equations with multiple variables that are solved together. They are fundamental in mathematics as they model situations where multiple conditions must be satisfied at the same time.
Solving simultaneous equations generally involves finding a unique set of values for the unknowns that make all the equations true. Gaussian elimination is one method to achieve this.
It provides a clear path to finding solutions and understanding the relationships between the variables.
Solving simultaneous equations generally involves finding a unique set of values for the unknowns that make all the equations true. Gaussian elimination is one method to achieve this.
- First, represent the system using an augmented matrix.
- Transform the matrix to upper triangular form.
- Use back substitution to find the variable values one by one.
It provides a clear path to finding solutions and understanding the relationships between the variables.
