Problem 49
Question
For Problems \(45-50\), change each augmented matrix of the system to reduced echelon form and then indicate the solutions of the system. \(\left(\begin{array}{r}x-2 y+4 z=9 \\ 2 x-4 y+8 z=3\end{array}\right)\)
Step-by-Step Solution
Verified Answer
The system has no solutions due to a contradiction in the augmented matrix.
1Step 1: Write the Augmented Matrix
First, convert the given system of equations into an augmented matrix.The system is:\[\begin{align*}x - 2y + 4z &= 9 \2x - 4y + 8z &= 3\end{align*}\]The corresponding augmented matrix is:\[\begin{bmatrix}1 & -2 & 4 & | & 9 \2 & -4 & 8 & | & 3\end{bmatrix}\]
2Step 2: Eliminate the First Column Below the Pivot
To get the first leading 1, we use the first row and eliminate the first entry of the second row.Subtract 2 times the first row from the second row:\[R2 = R2 - 2\cdot R1\]The new matrix becomes:\[\begin{bmatrix}1 & -2 & 4 & | & 9 \0 & 0 & 0 & | & -15\end{bmatrix}\]
3Step 3: Analyze the Second Row
The second row of the matrix is now \([0, 0, 0 | -15]\), which corresponds to the equation \(0 = -15\).This is a contradiction, indicating that the system has no solutions.
4Step 4: Conclude with No Solutions
Since we have a row that leads to a contradiction in the system \([0, 0, 0 | -15]\), it confirms there are no values of \(x\), \(y\), and \(z\) that satisfy both equations simultaneously. Therefore, the system of equations has no solutions.
Key Concepts
Understanding Augmented MatricesRecognizing No Solutions in a System of EquationsBasics of Linear Equations
Understanding Augmented Matrices
An augmented matrix is a key concept in solving systems of linear equations. It represents the coefficients and constant terms of the linear equations in a compact form. For example, when you have a system of equations, you can write this system as an augmented matrix by including all the coefficients and constants.
- The coefficients of the variables form the main part of the matrix.
- The constants are appended at the end, separated by a vertical bar to show they’re from the equations’ right-hand sides.
Recognizing No Solutions in a System of Equations
When dealing with systems of linear equations, it's crucial to determine if a solution exists, and if so, to find it. One key indicator of a system having no solutions occurs during the process of transforming the augmented matrix into reduced echelon form. This process involves performing row operations to simplify the matrix. As you progress, you might encounter a row that results in a scenario like this: \[0x + 0y + 0z = -15\] Here, the left side of the equation is completely zero, but the right side is a non-zero number. This implies a contradiction since 0 cannot equal -15. When such a contradiction appears, it indicates the system is inconsistent and has no solutions. In our example, after performing row operations on the original augmented matrix, we find ourselves with this contradiction, confirming there are no solutions.
Basics of Linear Equations
Linear equations are the foundational building blocks for solving systems like the one presented above. These equations take on a form similar to \[ ax + by + cz = d \] where \(a, b,\) and \(c\) are coefficients of the variables \(x, y,\) and \(z\), respectively, and \(d\) is the constant term.
When you have multiple linear equations, you can use an augmented matrix as a tool to solve them. Using techniques such as row reduction helps in translating these equations into a more manageable form, eventually finding solutions or identifying inconsistencies.
When you have multiple linear equations, you can use an augmented matrix as a tool to solve them. Using techniques such as row reduction helps in translating these equations into a more manageable form, eventually finding solutions or identifying inconsistencies.
- Linear equations are easy to manipulate using matrix operations.
- While solving, they are aimed at reducing complexity through elimination.
Other exercises in this chapter
Problem 48
For Problems \(45-50\), change each augmented matrix of the system to reduced echelon form and then indicate the solutions of the system. \(\left(\begin{array}{
View solution Problem 48
Solve each system by using either the substitution method or the elimination- by-addition method, whichever seems more appropriate. \(\left(\begin{array}{l}y=4
View solution Problem 50
Solve each system by using either the substitution method or the elimination- by-addition method, whichever seems more appropriate. \(\left(\begin{array}{c}y=\f
View solution Problem 51
Solve each system by using either the substitution method or the elimination- by-addition method, whichever seems more appropriate. \(\left(\begin{array}{l}\fra
View solution