Problem 25 - Precalculus

Question

Solve the following systems of linear equations using the techniques discussed in this section. Compare and contrast these techniques with those you used to solve the systems in the Exercises in Section 8.1 . $$ \left\\{\begin{aligned} 2 x-3 y+z &=-1 \\ 4 x-4 y+4 z &=-13 \\ 6 x-5 y+7 z &=-25 \end{aligned}\right. $$

Step-by-Step Solution

Answer

Infinitely many solutions with parameters: \(x = -\frac{35}{4} - 2z\), \(y = -\frac{11}{2} - z\), and \(z = z\). System is dependent.

Step 1: Choose a Method

The system of equations can be solved using several methods like substitution, elimination, or matrix (Gaussian elimination). Let's begin by using Gaussian elimination to solve this system.

Step 2: Write the Augmented Matrix

First, express the given system of equations as an augmented matrix:\[\begin{bmatrix} 2 & -3 & 1 & | & -1 \ 4 & -4 & 4 & | & -13 \ 6 & -5 & 7 & | & -25 \end{bmatrix} \]

Step 3: Row Operations to Simplify

Perform row operations to simplify the augmented matrix to row-echelon form. Let's first use row 1 to eliminate the first coefficient in rows 2 and 3:1. \( R2 = R2 - 2 \, (R1) \) (Subtracting twice the first row from the second row) New row 2: \([0, 2, 2, |, -11]\)2. \( R3 = R3 - 3 \, (R1) \) (Subtracting three times the first row from the third row) New row 3: \([0, 4, 4, |, -22]\)Augmented matrix becomes:\[\begin{bmatrix} 2 & -3 & 1 & | & -1 \ 0 & 2 & 2 & | & -11 \ 0 & 4 & 4 & | & -22 \end{bmatrix}\]

Step 4: Further Simplify Augmented Matrix

Next, simplify further to get the last matrix into row-echelon form. Use row 2 to eliminate the leading term in row 3:1. \( R3 = R3 - 2 \, (R2) \) (Subtracting twice the second row from the third row) New row 3: \([0, 0, 0, |, 0]\)Now the augmented matrix is:\[\begin{bmatrix} 2 & -3 & 1 & | & -1 \ 0 & 2 & 2 & | & -11 \ 0 & 0 & 0 & | & 0 \end{bmatrix}\]

Step 5: Back-Substitution

We have now a row-echelon matrix, but since the third row is zero, the system is dependent. Using back-substitution, express the variables:1. From row 2: \(2y + 2z = -11\) Simplify to get: \(y = -\frac{11}{2} - z\).2. From row 1: \(2x - 3y + z = -1\). Substitute \(y\) expression: \(2x + \frac{33}{2} + 3z + z = -1\), solve it for \(x\).

Step 6: Express Solutions of Variables

Solve from the simplified substitutions:1. From step 5, express \(x\) in terms of \(z\): \(2x = -1 - \frac{33}{2} - 4z\) gives \(x = -\frac{35}{4} - 2z\)2. From step 5, we already have \(y = -\frac{11}{2} - z\).Thus, the general solution is:\[x = -\frac{35}{4} - 2z, \quad y = -\frac{11}{2} - z, \quad z = z \]

Step 7: Solution Discussion

The system of equations has infinitely many solutions because the last row in the augmented matrix is all zeros, leading to a dependent system. Solutions can be expressed in terms of a parameter, \(z\).

Key Concepts

Augmented Matrix

In the process of solving systems of linear equations, an augmented matrix simplifies the equation-handling procedure. An augmented matrix is a concise way of writing a linear system by capturing both coefficients of variables and constant terms.

• Each row in the matrix corresponds to one equation from the system.
• The columns represent the coefficients of each variable, with an extra column representing the constant terms.
• This matrix form aids in applying Gaussian elimination effectively since it puts all the elements needed to solve the system into a single matrix.

For the given system, the matrix representation is constructed as follows: \[\begin{bmatrix} 2 & -3 & 1 & | & -1 \4 & -4 & 4 & | & -13 \6 & -5 & 7 & | & -25 \end{bmatrix}\]
This setup is essential before performing row operations to simplify and eventually solve for the variable values.

Row-Echelon Form

Converting an augmented matrix to row-echelon form (REF) is a critical step in solving linear systems using Gaussian elimination. The goal is to simplify the matrix to a form where solving for variables becomes more straightforward.

• In REF, each leading entry of a nonzero row is 1, and is the only nonzero entry in its column below it.
• Leading entries move strictly to the right as you move down the rows.
• Rows that are completely zero, if any, are moved to the bottom of the matrix.

The process involves performing row operations such as row swapping, scaling rows, or adding multiples of one row to another. For the given system, after the operations, the matrix looks like this: \[\begin{bmatrix} 2 & -3 & 1 & | & -1 \0 & 2 & 2 & | & -11 \0 & 0 & 0 & | & 0 \end{bmatrix}\]
With the matrix in this form, back-substitution becomes straightforward, allowing us to express solutions for each variable.

Dependent System

A dependent system arises when the equations in the system do not provide enough unique information to determine a unique solution. This generally results from one or more equations being a linear combination of the others.

• In a dependent system, at least one row in the row-echelon form is entirely zero, indicating redundancy in the equations.
• This redundancy means there aren't enough independent equations for each variable.
• It typically leads to the need for parametrization of solutions.

In the given system, the last row becomes all zeros, signifying dependency. Therefore, the system doesn't have a unique solution, and instead, one variable will serve as a parameter for the others.

Infinitely Many Solutions

When a system of linear equations has infinitely many solutions, it means that there are multiple solutions that satisfy all the equations. This scenario occurs in a dependent system and is characterized by having at least one free variable.

• These free variables can take any value, leading to different solutions each time they change.
• The solutions are usually expressed in terms of one or more parameters.
• Such parametrized solutions provide a clear description of the solution space.

For the system given, the variable \( z \) is a free parameter, and the solutions can be expressed as:\[x = -\frac{35}{4} - 2z, \quad y = -\frac{11}{2} - z, \quad z = z\]
This representation illustrates that by choosing any value for \( z \), corresponding values of \( x \) and \( y \) can be calculated, allowing for infinitely many valid solutions.