Problem 30 - A Graphical Approach to Precalculus with Limits

Question

Each augmented matrix is in row echelon form and represents a linear system. Use back-substitution to solve the system if possible. $$\left[\begin{array}{rrr|r} 1 & 0 & -4 & \frac{3}{4} \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & -3 \end{array}\right]$$

Step-by-Step Solution

Answer

The system has no solutions due to a row indicating a contradiction.

Step 1: Identify the System of Equations

The augmented matrix corresponds to the following system of linear equations: 1. \( x - 4z = \frac{3}{4} \) 2. \( y + 2z = 1 \) 3. The third row, \( 0 = -3 \), represents a contradiction, implying no solutions to this system.

Step 2: Analyze the Third Row

Observe that the third row shows an equation \( 0 = -3 \). This is a contradiction, indicating that the system has no solutions.

Step 3: Conclude the Solution

Since the third row results in \( 0 = -3 \), which is a false statement, the system is inconsistent and does not have any solutions.

Key Concepts

augmented matrix

An augmented matrix is a powerful tool used to represent and solve systems of linear equations. Imagine having a set of equations that you want to solve simultaneously. By organizing these equations into a matrix, it becomes easier to apply mathematical techniques to find solutions or identify if a solution exists.
In the context of our example, the augmented matrix was written as follows:

• The numbers to the left of the vertical line represent the coefficients of the variables in the system.
• The numbers to the right of the vertical line stand for the constants from each equation.

This visual representation simplifies operations like row reductions and allows for easy interpretation of the system's structure, paving the way to solving or analyzing the system.

back-substitution

Back-substitution is a method used to find solutions to a linear system written in triangular form, typically after reducing the augmented matrix to a form where you can easily solve for each variable step-by-step. However, in our current exercise, back-substitution wasn't possible due to the third row highlighting a contradiction.
The process generally involves solving the equations starting from the bottom up and substituting up through the system:

• First, solve the equation represented by the last row for its corresponding variable.
• Then, proceed upward, using the already found values to solve for the upper variables.

However, since our matrix indicates an inconsistency, the step of back-substitution cannot derive a solution, as the system contradicts itself.

row echelon form

Row echelon form is achieved when an augmented matrix has been manipulated to a state where each leading entry of a row is to the right of the leading entry of the row above it. This form facilitates the process of solving linear systems, as it simplifies the task of recognizing which variables are dependent or independent.
For the given exercise, the matrix is already in row echelon form:

• The first two rows form a staircase pattern where the leading coefficients (1's) are positioned top to bottom.
• Each leading 1 is to the right of the one above it.
• All entries below the leading 1's are zero.

This structure normally aids the use of back-substitution, but an inconsistency in our system (third row) indicates no need for further computation.

inconsistent system

An inconsistent system refers to a set of equations that does not have any solution. This arises when the mathematical depiction of the system shows contradictions, meaning the equations clash against one another during the solving process.
In the studied problem, the third row of the matrix is particularly revealing:

• The third row conveys the equation \(0 = -3\), a clear contradiction.
• Such a situation tells us that no value for the variables will satisfy all equations simultaneously.

When a system is inconsistent, it implies there's a flaw in compatibility between the equations. Identifying this inconsistency at an early stage is crucial as it helps avoid unnecessary calculations and one immediately knows that there are no solutions to pursue.