Problem 29
Question
Each augmented matrix is in row echelon form and represents a linear system. Use back-substitution to solve the system if possible. $$\left[\begin{array}{rrr|r} 1 & 2 & 1 & -3 \\ 0 & 1 & -3 & \frac{1}{2} \\ 0 & 0 & 0 & 4 \end{array}\right]$$
Step-by-Step Solution
Verified Answer
The system is inconsistent as the equation 0 = 4 cannot be true, so there is no solution.
1Step 1: Analyze the Augmented Matrix
First, examine the given augmented matrix. The matrix represents the following system of equations: \[ \begin{align*} x + 2y + z &= -3 \ y - 3z &= \frac{1}{2} \ 0 &= 4 \end{align*} \] Notice that the third row, \(0 = 4\), is inconsistent, which indicates that there is no solution to this system of equations.
2Step 2: Identify the Issue
The third row, \(0 = 4\), is a contradiction because zero cannot equal a non-zero number. This inconsistency implies that the system does not have any solutions. This means we cannot proceed with back-substitution since there is no way to satisfy all equations simultaneously.
Key Concepts
Augmented MatrixRow Echelon FormBack-Substitution
Augmented Matrix
When dealing with systems of linear equations, an augmented matrix serves as a concise way to represent the equations and their constants. It puts all the information side by side. The structure of an augmented matrix embeds the coefficients of the variables and the constants from the equations in a single matrix.
For instance, consider the matrix provided in the exercise: \[\begin{bmatrix}1 & 2 & 1 & | & -3 \0 & 1 & -3 & | & \frac{1}{2} \0 & 0 & 0 & | & 4\end{bmatrix}\]This format allows us to easily manage and manipulate the system during solving processes like Gaussian elimination.
For instance, consider the matrix provided in the exercise: \[\begin{bmatrix}1 & 2 & 1 & | & -3 \0 & 1 & -3 & | & \frac{1}{2} \0 & 0 & 0 & | & 4\end{bmatrix}\]This format allows us to easily manage and manipulate the system during solving processes like Gaussian elimination.
- The left part (before the vertical line) contains the coefficients of the variables \(x\), \(y\), and \(z\).
- The right part (after the vertical line) lists the constants of the equations.
Row Echelon Form
Row echelon form (REF) is an essential concept in linear algebra, often used as a stepping stone for solving systems of equations. It simplifies the augmented matrix in a way that makes it easier to solve or analyze for consistency. REF is characterized by a set of specific rules.
\[\begin{bmatrix}1 & 2 & 1 & | & -3 \0 & 1 & -3 & | & \frac{1}{2} \0 & 0 & 0 & | & 4\end{bmatrix}\]
This form streamlines the process of back-substitution if the matrix reflection shows a consistent and solvable system. However, a row like \(0=4\) signifies inconsistency, so we can immediately identify that no solution exists without proceeding to the final solving phase.
- Each leading coefficient (the first non-zero number from the left, in each row) is to the right of the leading coefficient in the row above.
- All rows containing all zero values are at the bottom of the matrix.
- The leading coefficient in each non-zero row is 1.
\[\begin{bmatrix}1 & 2 & 1 & | & -3 \0 & 1 & -3 & | & \frac{1}{2} \0 & 0 & 0 & | & 4\end{bmatrix}\]
This form streamlines the process of back-substitution if the matrix reflection shows a consistent and solvable system. However, a row like \(0=4\) signifies inconsistency, so we can immediately identify that no solution exists without proceeding to the final solving phase.
Back-Substitution
Back-substitution is the process used to find the solution of a system of linear equations after the augmented matrix has been simplified to row echelon form. It essentially 'works backwards' from the bottom-most equation upwards. For consistent systems, this means solving the last equation first and then substituting those values into the equations above to find the other variable values.
In our scenario, although the matrix was in REF, the presence of the row "\(0 = 4\)" means we can't perform back-substitution as the system is inconsistent. That row suggests a contradiction.
In our scenario, although the matrix was in REF, the presence of the row "\(0 = 4\)" means we can't perform back-substitution as the system is inconsistent. That row suggests a contradiction.
- Normally, in performing back-substitution, you use the last non-zero row to find one variable.
- Substitute this found value into the row above to solve for another variable.
- Repeat this process until all variables have known values.
Other exercises in this chapter
Problem 29
Graph each inequality. $$y \leq \frac{1}{x}$$
View solution Problem 29
Solve each determinant equation for \(x\). $$\operatorname{det}\left[\begin{array}{ll}2 x & x \\\11 & x\end{array}\right]=6$$
View solution Problem 29
Find the partial fraction decomposition for each rational expression. $$\frac{1}{x(2 x+1)\left(3 x^{2}+4\right)}$$
View solution Problem 29
For each matrix, find \(A^{-1}\) if it exists. $$A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 5 & 10 & 5 \\ 3 & 6 & 3 \end{array}\right]$$
View solution