Problem 7
Question
Given that the augmented matrix in row-reduced form is equivalent to the augmented matrix of a system of linear equations, (a) determine whether the system has a solution and (b) find the solution or solutions to the system, if they exist. \(\left[\begin{array}{llll|l}1 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 0 & 1\end{array}\right]\)
Step-by-Step Solution
Verified Answer
The given row-reduced augmented matrix represents an inconsistent system of linear equations, as the last row indicates an impossible equation: \(0x + 0y + 0z + 0w = 1\). Therefore, the system does not have a solution.
1Step 1: Identify the Last Row
The given matrix has 4 rows, and we need to look at the last row to determine if there's a solution to the system of linear equations. The last row of the matrix is:
\[0 ~ 0 ~ 0 ~ 0 ~|~ 1\]
2Step 2: Check for Inconsistency
If the last row from the given matrix has all zeros on the left but a non-zero value (1) on the right side of the augmented part, it means that the system is inconsistent and there is no solution.
In this case, the last row indicates an inconsistent equation: \(0x + 0y + 0z + 0w = 1\), which is impossible.
3Step 3: Determine Solution Existence
Since the last row of the matrix represents an inconsistent equation, the given system of linear equations does not have a solution.
#Phase 2: Find the solution(s) to the system#
4Step 4: No Solution
As we've concluded in Phase 1, the given system of linear equations does not have a solution, so there are no solutions to find.
In summary, the augmented row-reduced matrix represents a system of linear equations with no solution.
Key Concepts
Row-Reduced Echelon FormInconsistent SystemAugmented MatrixSystems of Linear Equations
Row-Reduced Echelon Form
The row-reduced echelon form (RREF) is a crucial topic in linear algebra. This form of a matrix simplifies solving systems of linear equations. An augmented matrix is considered to be in row-reduced echelon form if it meets a few specific criteria:
- Each leading non-zero entry of a row, known as a leading 1, is the only non-zero entry in its column.
- Leading 1s move to the right as you move down the rows.
- All rows consisting entirely of zeros are at the bottom of the matrix.
Inconsistent System
An inconsistent system is a set of equations that has no solution. This inconsistency often becomes apparent when the system is converted into row-reduced echelon form. Inconsistent systems are identified when there is a row that has the following form:
- All zero coefficients on the left side of the augmented matrix.
- A non-zero constant on the right side, after the augmentation line.
Augmented Matrix
An augmented matrix is an essential representation in linear algebra that helps solve systems of linear equations. It is constructed by appending the constants of the equations to the right of their coefficients. Simply put:
- A normal matrix represents the coefficients of variables in the system.
- The augmented matrix includes an additional column that stands for the constant terms from the equations' right-hand sides.
Systems of Linear Equations
Systems of linear equations involve finding values for unknown variables that satisfy multiple equations simultaneously. Each equation represents a condition that must be true for those variable values. Key characteristics of these systems include:
- They can be consistent (having at least one solution) or inconsistent (no solutions exist).
- They can also be independent (exactly one solution) or dependent (infinitely many solutions).
Other exercises in this chapter
Problem 7
Compute the indicated products. \(\left[\begin{array}{ll}1 & 2 \\ 3 & 0\end{array}\right]\left[\begin{array}{r}1 \\ -1\end{array}\right]\)
View solution Problem 7
Refer to the following matrices: \(A=\left[\begin{array}{rr}-1 & 2 \\ 3 & -2 \\ 4 & 0\end{array}\right] \quad B=\left[\begin{array}{rr}2 & 4 \\ 3 & 1 \\ -2 & 2\
View solution Problem 7
Write the system of equations corresponding to each augmented matrix. \(\left[\begin{array}{rrr|r}1 & 3 & 2 & 4 \\ 2 & 0 & 0 & 5 \\ 3 & -3 & 2 & 6\end{array}\ri
View solution Problem 7
Determine whether each system of linear equations has (a) one and only one solution, (b) infinitely many solutions, or (c) no solution. Find all solutions whene
View solution