Problem 10
Question
Solve each of the following systems. If the solution set is \(\varnothing\) or if it contains infinitely many solutions, then so indicate. $$ \left(\begin{array}{r} x+y-z=2 \\ 3 x-4 y+2 z=5 \\ 2 x+2 y-2 z=7 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
The system has no solution; the solution set is \(\varnothing\).
1Step 1: Write the System in Matrix Form
Represent the system of equations as an augmented matrix:\[\begin{bmatrix}1 & 1 & -1 & | & 2 \3 & -4 & 2 & | & 5 \2 & 2 & -2 & | & 7\end{bmatrix}\]
2Step 2: Apply Gaussian Elimination
Transform the augmented matrix to row-echelon form (REF):Step 2a: Use the first row to eliminate the first column entries in rows 2 and 3.- Row 2: \(R_2 - 3R_1 \rightarrow R_2\) \[\begin{bmatrix}0 & -7 & 5 & | & -1\end{bmatrix}\]- Row 3: \(R_3 - 2R_1 \rightarrow R_3\) \[\begin{bmatrix}0 & 0 & 0 & | & 3\end{bmatrix}\]Now the matrix looks like this:\[\begin{bmatrix}1 & 1 & -1 & | & 2 \0 & -7 & 5 & | & -1 \0 & 0 & 0 & | & 3\end{bmatrix}\]
3Step 3: Analyze Row Echelon Form
In the row-echelon form, the third row translates to the equation \(0x + 0y + 0z = 3\), which is an inconsistency because it equates to \(0 = 3\). This indicates that the system of equations is inconsistent.
4Step 4: Determine the Solution Set
Since we found an inconsistency in Step 3, the solution set for the system is \(\varnothing\), meaning there are no solutions.
Key Concepts
Gaussian eliminationInconsistent systemAugmented matrix
Gaussian elimination
Gaussian elimination is a fundamental method to solve systems of linear equations. It uses a series of operations to transform a system into a simpler form, making it easier to find solutions. When working with Gaussian elimination, we focus on manipulating an augmented matrix until it reaches a form called the row-echelon form (REF). This involves using the leading entry (or pivot) of each row to make the elements below it zero.
In the example given, Gaussian elimination starts with arranging the system of equations into an augmented matrix. This matrix encapsulates all the coefficients of the variables and their corresponding constant terms. Through careful row operations, the augmented matrix is manipulated to achieve REF:
- Swap, multiply, or add rows to make leading entries of each row below the leading entry of the next row zero. - Continue these steps until each leading entry has zeros beneath it.
The goal of REF is to easily back-substitute to find solutions. However, if an inconsistency arises, as we see in this case, it indicates the system has no solutions.
In the example given, Gaussian elimination starts with arranging the system of equations into an augmented matrix. This matrix encapsulates all the coefficients of the variables and their corresponding constant terms. Through careful row operations, the augmented matrix is manipulated to achieve REF:
- Swap, multiply, or add rows to make leading entries of each row below the leading entry of the next row zero. - Continue these steps until each leading entry has zeros beneath it.
The goal of REF is to easily back-substitute to find solutions. However, if an inconsistency arises, as we see in this case, it indicates the system has no solutions.
Inconsistent system
An inconsistent system of linear equations means there are no solutions that satisfy all the equations simultaneously. This occurs when there is a contradiction within the system.
For the given problem, once the system is transformed using Gaussian elimination, the presence of a row like \[0x + 0y + 0z = 3\]indicates a contradiction, as it simplifies to \[0 = 3\]. Such an equation is clearly impossible, since zero can never equal three.
An inconsistent system often displays this type of contradictory row in its row-echelon form. When solving, if you encounter such a row, you can immediately conclude the system has no solutions. This is why an inconsistent system is represented by the empty set, denoted as \(\varnothing\).
For the given problem, once the system is transformed using Gaussian elimination, the presence of a row like \[0x + 0y + 0z = 3\]indicates a contradiction, as it simplifies to \[0 = 3\]. Such an equation is clearly impossible, since zero can never equal three.
An inconsistent system often displays this type of contradictory row in its row-echelon form. When solving, if you encounter such a row, you can immediately conclude the system has no solutions. This is why an inconsistent system is represented by the empty set, denoted as \(\varnothing\).
Augmented matrix
An augmented matrix is a vital tool in solving systems of linear equations. It simplifies manipulation of the equations by bundling the coefficients together and aligning the constant terms on the right side, separated by a partition.
In the provided example, the system of equations is represented as \[\begin{bmatrix}1 & 1 & -1 & | & 2 \ 3 & -4 & 2 & | & 5 \ 2 & 2 & -2 & | & 7 \end{bmatrix}\]The left part of the matrix contains coefficients from the variables \(x, y,\) and \(z\). The right part, after the vertical line, includes the constant terms.
The benefit of arranging a system into an augmented matrix is that it lays the foundation for applying matrix operations like row swapping and combinations, crucial in Gaussian elimination. It helps to visually track and manage equations during transformation until the system reaches an easily interpretable form.
In the provided example, the system of equations is represented as \[\begin{bmatrix}1 & 1 & -1 & | & 2 \ 3 & -4 & 2 & | & 5 \ 2 & 2 & -2 & | & 7 \end{bmatrix}\]The left part of the matrix contains coefficients from the variables \(x, y,\) and \(z\). The right part, after the vertical line, includes the constant terms.
The benefit of arranging a system into an augmented matrix is that it lays the foundation for applying matrix operations like row swapping and combinations, crucial in Gaussian elimination. It helps to visually track and manage equations during transformation until the system reaches an easily interpretable form.
Other exercises in this chapter
Problem 10
For Problems \(1-10\), use expansion by minors to evaluate each determinant. (Objective 1) $$ \left|\begin{array}{rrr} -5 & 2 & 6 \\ 1 & -1 & 3 \\ 4 & -2 & -4 \
View solution Problem 10
Evaluate each of the following determinants. $$ \left|\begin{array}{rr} -2 & 4 \\ 9 & -7 \end{array}\right| $$
View solution Problem 10
For Problems \(1-18\), use the elimination-by-addition method to solve each system. (Objective 1 ) $$ \left(\begin{array}{l} 4 x+3 y=-4 \\ 3 x-7 y=34 \end{array
View solution Problem 10
For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{l} y=3 x-5 \\ 2 x+3 y=6 \end{array}\right) $$
View solution