Problem 14 - Precalculus

Question

In Exercises \(9-26\), put each system of linear equations into triangular form and solve the system if poesible. Classify each system as consistent independent, consistent dependent, or inconsistent. $$ \left\\{\begin{aligned} x-2 y+3 z &=7 \\ -3 x+y+2 z &=-5 \\ 2 x+2 y+z &=3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is consistent and independent.

Step 1: Convert to Matrix Form

First, we represent the system of equations in matrix form, which is easier for transformation. The equations in matrix form are: \[\begin{bmatrix}1 & -2 & 3 & | & 7 \-3 & 1 & 2 & | & -5 \2 & 2 & 1 & | & 3\end{bmatrix}.\]

Step 2: Apply Row Reduction to Matrix

We use the Gauss-Jordan elimination process to put the matrix in upper triangular form. Start with the leading coefficient (pivot) in the first row, and make elements below this pivot zero. We first make the first element of Row 2 and Row 3 zero by using Row 1:- For Row 2: Multiply Row 1 by 3 and add to Row 2.- For Row 3: Multiply Row 1 by -2 and add to Row 3.The updated matrix is:\[\begin{bmatrix}1 & -2 & 3 & | & 7 \0 & -5 & 11 & | & 16 \0 & 6 & -5 & | & -11\end{bmatrix}.\]

Step 3: Eliminate Below Next Pivot

Focus on the second pivot, \(-5\), in Row 2 and make elements below it zero:- Multiply Row 2 by 6 then add to Row 3 after multiplying Row 3 by 5. This step ensures the second entry of Row 3 becomes zero to achieve triangular form. After adjustments, the matrix becomes:\[\begin{bmatrix}1 & -2 & 3 & | & 7 \0 & -5 & 11 & | & 16 \0 & 0 & 31 & | & 59\end{bmatrix}.\]

Step 4: Solve the System Starting from the Bottom

Now, we solve the system using back substitution. Start with the last equation (third row), which gives:\[31z = 59\]Solve for \(z\):\[z = \frac{59}{31}.\]

Step 5: Back-Substitute to Find Other Variables

Substitute \(z = \frac{59}{31}\) back into the second equation to find \(y\):\[0y - 5y + 11\left(\frac{59}{31}\right) = 16\]Simplify and solve for \(y\):\[-5y + \frac{649}{31} = 16\]\[-5y = 16 - \frac{649}{31}\]Solve \(y\) carefully. Repeat substitution for first equation to find \(x\).

Step 6: Solve for Consistency and Dependency

The system in triangular form gives a unique solution when solved step-by-step with unique values for \(x,y,z\). Therefore, this system of equations is consistent and independent.

Key Concepts

System of Linear Equations

In mathematics, a system of linear equations is a collection of linear equations involving the same set of variables. These equations represent straight lines in a coordinate system, or planes if there are three variables, and so on. The solution to a system of linear equations is the set of values for the variables that makes all the equations true simultaneously.
For example, consider the system given in the original exercise:

• Equation 1: \( x - 2y + 3z = 7 \)
• Equation 2: \( -3x + y + 2z = -5 \)
• Equation 3: \( 2x + 2y + z = 3 \)

These equations involve three unknowns: \( x, y, \text{ and } z \), and we need to find a combination of these unknown values that satisfies all three equations.
Solving systems of linear equations can be done using several methods, such as substitution, elimination, or matrix operations like Gauss-Jordan elimination.

Matrix Representation

Using matrix representation for a system of linear equations is one of the most efficient and systematic ways to solve these problems. In matrix form, the system of equations is expressed as a compact and structured array of numbers.
To convert our given system to matrix form, we represent the system as an augmented matrix:

• The first row represents the coefficients of the variables in the first equation.
• The second row corresponds to the second equation.
• The numbers to the right of the vertical line are the constants on the other side of the equations.

Here's how the original system translates:\[\begin{bmatrix} 1 & -2 & 3 & | & 7 \-3 & 1 & 2 & | & -5 \2 & 2 & 1 & | & 3 \\end{bmatrix}\]
The matrix succinctly contains all the necessary information from the system of equations, making them ready for a variety of matrix solution techniques.

Gauss-Jordan Elimination

Gauss-Jordan elimination is a step-by-step process for transforming a matrix into a specific form where solving the system becomes straightforward. This method involves row operations to systematically eliminate variables and reach a triangular form.
The steps include:

• Select a leading coefficient (pivot) in each row.
• Use the pivot to eliminate elements below it in the same column, making them zero.
• Continue this process for each row until the matrix reaches a situation where back substitution can be used.

In our exercise, Gauss-Jordan elimination transformed the matrix to:\[\begin{bmatrix} 1 & -2 & 3 & | & 7 \0 & -5 & 11 & | & 16 \0 & 0 & 31 & | & 59 \\end{bmatrix}\]
This triangular form allows us to solve for one variable at a time, beginning with the last row and substituting back through the previous rows.

Consistent Independent System

After applying the Gauss-Jordan elimination method, we can determine whether a system of equations is consistent and independent. A consistent system has at least one solution, meaning the equations intersect at a common point or points.
If a system has exactly one unique solution, it is said to be independent. In triangular form, this is indicated by having non-zero pivots leading to unique solutions for each variable.
In the exercise, we found a unique solution for \( z \) from the third row, which then allowed us to find unique values for \( y \) and \( x \). The fact that all the variables have unique solutions suggests that this system is consistent and independent.
A consistent independent system is desirable when one needs a precise and unique solution for real-world problems.