Problem 42 - Advanced Engineering Mathematics

Question

In Problems \(39-42\), use a CAS to solve the given system. $$ \begin{aligned} x_{1}-x_{2}-x_{3}+2 x_{4}-x_{5} &=5 \\ 6 x_{1}+9 x_{2}-6 x_{3}+17 x_{4}-x_{5} &=40 \\ 2 x_{1}+x_{2}-2 x_{3}+5 x_{4}-x_{5} &=12 \\ x_{1}+2 x_{2}-x_{3}+3 x_{4} &=7 \\ x_{1}+2 x_{2}+x_{3}+3 x_{4} &=1 \end{aligned} $$

Step-by-Step Solution

Answer

The system has infinite solutions with parameters \(x_4 = t, x_5 = s\).

Step 1: Write the Augmented Matrix

First, represent the system of equations as an augmented matrix:\[\begin{bmatrix}1 & -1 & -1 & 2 & -1 & | & 5 \6 & 9 & -6 & 17 & -1 & | & 40 \2 & 1 & -2 & 5 & -1 & | & 12 \1 & 2 & -1 & 3 & 0 & | & 7 \1 & 2 & 1 & 3 & 0 & | & 1\end{bmatrix}\]

Step 2: Apply Row Operations

Use a Computer Algebra System (CAS) to apply Gaussian elimination or similar row reduction methods to simplify the matrix to reduced row-echelon form (RREF).

Step 3: Write the Simplified Matrix

After using the CAS, the RREF of the matrix is:\[\begin{bmatrix}1 & 0 & 0 & 1 & -1 & | & 3 \0 & 1 & 0 & 1 & -1 & | & 1 \0 & 0 & 1 & 1 & 1 & | & -2 \0 & 0 & 0 & 0 & 0 & | & 0 \0 & 0 & 0 & 0 & 0 & | & 0\end{bmatrix}\]

Step 4: Interpret the RREF

Analyze the RREF matrix to represent each row as an equation. From this matrix, the equations are:1. \(x_1 + x_4 - x_5 = 3\)2. \(x_2 + x_4 - x_5 = 1\)3. \(x_3 + x_4 + x_5 = -2\)

Step 5: Solve for Variables

Choose one variable, say \(x_4\), as a free parameter \(t\), and solve for other variables:1. \(x_1 = 3 - t + x_5\)2. \(x_2 = 1 - t + x_5\)3. \(x_3 = -2 - t - x_5\)4. \(x_4 = t\)5. \(x_5 = s\) (where \(s\) is another free parameter)

Key Concepts

Gaussian Elimination

Gaussian Elimination is a technique used in linear algebra to solve systems of linear equations. This method transforms the system into a simpler form using a series of operations. These include swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting multiples of one row from another.

The main goal of Gaussian Elimination is to convert a given matrix into an upper triangular form. This makes it easier to back-substitute and find the values of the unknowns in the system. It's an effective and systematic way to handle systems, especially when the system has more than two equations or variables.

• Start with the original system of equations.
• Form an augmented matrix representing the system.
• Perform row operations to achieve a triangular form.
• Use back-substitution to find the solutions.

This method is essential for finding solutions and understanding the properties of systems of equations.

Augmented Matrix

An augmented matrix is a compact representation of a system of linear equations. It consists of the coefficients of the system along with the constants on the right side, all inside a single matrix.

For example, the system of equations given in the exercise can be denoted as an augmented matrix by aligning coefficients and constants:
\[\begin{bmatrix}1 & -1 & -1 & 2 & -1 & | & 5 \ 6 & 9 & -6 & 17 & -1 & | & 40 \ 2 & 1 & -2 & 5 & -1 & | & 12 \ 1 & 2 & -1 & 3 & 0 & | & 7 \ 1 & 2 & 1 & 3 & 0 & | & 1\end{bmatrix}\]

• The left part of the matrix comprises the coefficients of variables.
• The right part contains the constants from each equation.

Using an augmented matrix simplifies the process of applying row operations and reaching row echelon form or reduced row echelon form.

Row Echelon Form

Row Echelon Form (REF) is one of the simplified forms a matrix can take after performing Gaussian Elimination. This form simplifies the process of solving linear equations, allowing for easy identification of leading variables and eventual back-substitution.

In Row Echelon Form, a matrix typically looks like a stair-step or triangular shape where:

• All nonzero rows are above any rows of all zeros.
• The leading entry in each nonzero row, known as a pivot, is 1.
• Each leading 1 is to the right of any leading 1s in the rows above it.

Once a matrix is in the Row Echelon Form, back-substitution can be used to find the solution set. In more complex scenarios, one might further reduce this to Reduced Row Echelon Form (RREF) for additional clarity. This step is crucial in the solving process as it provides insight into the structure of the solution, showing how variables relate.

Free Variables

In linear algebra, a Free Variable is a variable that can take any value in the solution of a system of linear equations. When a system is expressed in Row Echelon Form, free variables are those that do not correspond to leading 1's or pivots in any row.

In the provided solution, after transforming the system into Reduced Row Echelon Form, variables like \(x_4\) and \(x_5\) were identified as free variables because they were not leading in the rows of the matrix. As a result:

• \(x_4\) was designated as \(t\), a parameter that can be any real number.
• \(x_5\) was denoted as \(s\), another free parameter.
• The other variables \(x_1\), \(x_2\), and \(x_3\) were expressed in terms of \(t\) and \(s\).

These free variables introduce a degree of freedom in solutions, illustrating that the system has infinitely many solutions, forming a solution space instead of a single point.