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Magazine Chapter 14: Problem 4
Solve the system of equations. $$\begin{aligned} x+y+z &=6 \\ 2 x-y-z &=-3 \\ x-2 y+3 z &=6 \end{aligned}$$
Short Answer
Expert verified
The solution to the system of equations is \(x = 1\), \(y = 2\), and \(z = 3\).
Step by step solution
01
Setting up the augmented matrix
:
First, we will represent the system of equations as an augmented matrix. This will make it easier to do the required operations for solving the system.
\( \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 2 & -1 & -1 & | & -3 \\ 1 & -2 & 3 & | & 6 \end{bmatrix} \)
02
Gaussian elimination
:
We'll apply the Gaussian elimination method. The goal is to create zeros below the main diagonal, transforming the matrix into an upper triangular matrix.
1. Subtract equation (1) from equation (3) to eliminate x from equation (3).
\( \text{New Equation 3} = \text{Equation 3} - \text{Equation 1} \)
\( \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 2 & -1 & -1 & | & -3 \\ 0 & -3 & 2 & | & 0 \end{bmatrix} \)
2. Subtract twice equation (1) from equation (2) to eliminate x from equation (2).
\( \text{New Equation 2} = \text{Equation 2} - 2 (\text{Equation 1}) \)
\( \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & -3 & -3 & | & -15 \\ 0 & -3 & 2 & | & 0 \end{bmatrix} \)
3. Divide equation (2) by -3 to simplify.
\( \text{New Equation 2} = \frac{1}{-3} (\text{Equation 2}) \)
\( \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 1 & | & 5 \\ 0 & -3 & 2 & | & 0 \end{bmatrix} \)
4. Add 3 times equation (2) to equation (3) to eliminate y from equation (3).
\( \text{New Equation 3} = \text{Equation 3} + 3 (\text{Equation 2}) \)
\( \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 1 & | & 5 \\ 0 & 0 & 5 & | & 15 \end{bmatrix} \)
03
Solve for z, y, and x
:
Now that we have an upper triangular matrix, we can solve for z, y, and x using back-substitution.
1. Solve for z:
\(5z = 15 \Rightarrow z = 3\)
2. Substitute the value of z into equation (2) to solve for y:
\(y + z = 5 \Rightarrow y = 5 - 3 = 2\)
3. Substitute the values of y and z into equation (1) to solve for x:
\(x + y + z = 6 \Rightarrow x = 6 - 2 - 3 = 1\)
The solution to the system of equations is x = 1, y = 2, and z = 3.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gaussian elimination
Gaussian elimination is a systematic method used to solve systems of linear equations. It focuses on transforming the system into a simpler form that makes it easier to solve.
In our original exercise, we apply Gaussian elimination by eliminating variables step-by-step. Our first task is to set up the augmented matrix of the system, where each row represents an equation and each column represents the coefficients of the variables, including the constants on the right side. Then, we aim to create zeros below the leading coefficients (or pivots), "stepping down" the matrix. Each step involving arithmetic with the rows is crucial to gradually transforming the matrix into an easier form to solve. The process continues until the matrix reaches an upper triangular form, where every number below each pivot is zero.
- **Goal:** Transform the original matrix into an upper triangular matrix.
- **Procedure:** Use elementary row operations like row swapping, multiplying rows, and adding or subtracting rows.
In our original exercise, we apply Gaussian elimination by eliminating variables step-by-step. Our first task is to set up the augmented matrix of the system, where each row represents an equation and each column represents the coefficients of the variables, including the constants on the right side. Then, we aim to create zeros below the leading coefficients (or pivots), "stepping down" the matrix. Each step involving arithmetic with the rows is crucial to gradually transforming the matrix into an easier form to solve. The process continues until the matrix reaches an upper triangular form, where every number below each pivot is zero.
Augmented matrix
An augmented matrix is a vital representation that combines the coefficient matrix of a system of equations with the constants on the right side into a single matrix. This approach simplifies the operations needed to solve the system.
In our task, we start with an augmented matrix representing our system of equations:\[\begin{bmatrix} 1 & 1 & 1 & | & 6 \2 & -1 & -1 & | & -3 \1 & -2 & 3 & | & 6 \end{bmatrix}\]
This matrix correlates each row with an equation. The vertical line separates the coefficients of the variables (left side) from the constants (right side). By working directly with this matrix, operations become more visual and systematic, aiding the transition to simpler forms necessary for solving the equations.
- **Format:** The augmented matrix adds an extra column to the end of the coefficient matrix.
- **Purpose:** It allows us to apply Gaussian elimination without needing to continuously refer back to the original equations.
In our task, we start with an augmented matrix representing our system of equations:\[\begin{bmatrix} 1 & 1 & 1 & | & 6 \2 & -1 & -1 & | & -3 \1 & -2 & 3 & | & 6 \end{bmatrix}\]
This matrix correlates each row with an equation. The vertical line separates the coefficients of the variables (left side) from the constants (right side). By working directly with this matrix, operations become more visual and systematic, aiding the transition to simpler forms necessary for solving the equations.
Back-substitution
Back-substitution is the final step in the process of solving a system of equations using Gaussian elimination. It occurs when an upper triangular matrix is obtained, making it straightforward to find the solutions.
Once we reach an upper triangular matrix like:\[\begin{bmatrix} 1 & 1 & 1 & | & 6 \0 & 1 & 1 & | & 5 \0 & 0 & 5 & | & 15 \end{bmatrix}\]
The last equation instantly gives the value for \(z\) since it's the only variable left in its row. Solving for \(z\), we find \(z = 3\). Moving upwards, we substitute \(z\) into the second row to determine \(y\), yielding \(y = 2\). Finally, substitute both \(y\) and \(z\) into the first row to find \(x = 1\).
Back-substitution is efficient and successfully untangles the variables, providing the specific solutions for each variable in the system.
- **Objective:** Solve for the variables starting from the last row upward.
- **Method:** Begin with the last variable, typically isolated in the last equation.
Once we reach an upper triangular matrix like:\[\begin{bmatrix} 1 & 1 & 1 & | & 6 \0 & 1 & 1 & | & 5 \0 & 0 & 5 & | & 15 \end{bmatrix}\]
The last equation instantly gives the value for \(z\) since it's the only variable left in its row. Solving for \(z\), we find \(z = 3\). Moving upwards, we substitute \(z\) into the second row to determine \(y\), yielding \(y = 2\). Finally, substitute both \(y\) and \(z\) into the first row to find \(x = 1\).
Back-substitution is efficient and successfully untangles the variables, providing the specific solutions for each variable in the system.
