Problem 12 - Advanced Engineering Mathematics

Question

Use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. \(x_{1}-x_{2}-2 x_{3}=0\) \(2 x_{1}+4 x_{2}+5 x_{3}=0\) \(6 x_{1}-\quad 3 x_{3}=0\)

Step-by-Step Solution

Answer

The system has infinitely many solutions parameterized by a free variable \(t\).

Step 1: Write the augmented matrix

First, express the given system of equations as an augmented matrix. The system of equations can be written in matrix form as follows:\[\begin{bmatrix}1 & -1 & -2 & | & 0 \2 & 4 & 5 & | & 0 \6 & 0 & -3 & | & 0 \\end{bmatrix}\]

Step 2: Begin Row Reduction

The first step in Gaussian elimination is to make the first pivot element (top-left corner) 1. In this case, it is already 1. Now, eliminate the elements below this pivot to create zeros. Replace row 2 (R_2) with R_2 - 2R_1 and row 3 (R_3) with R_3 - 6R_1:\[\begin{bmatrix}1 & -1 & -2 & | & 0 \0 & 6 & 9 & | & 0 \0 & 6 & 9 & | & 0 \\end{bmatrix}\]

Step 3: Simplify and Create Zero Below Pivot

Now, simplify row 2 by dividing it by 6 to make the pivot element 1, which results in:\[\begin{bmatrix}1 & -1 & -2 & | & 0 \0 & 1 & rac{3}{2} & | & 0 \0 & 6 & 9 & | & 0 \\end{bmatrix}\]Next, eliminate the element below this pivot (in the third row, second column) by subtracting 6 times row 2 from row 3: R_3 - 6R_2.

Step 4: Continue Row Reduction

Subtracting 6 times row 2 from row 3 gives us:\[\begin{bmatrix}1 & -1 & -2 & | & 0 \0 & 1 & rac{3}{2} & | & 0 \0 & 0 & 0 & | & 0 \\end{bmatrix}\]Now, our reduced matrix shows that the third row is all zeros, which implies there are infinitely many solutions.

Step 5: Parameterize the Solution

Since the third row is entirely zero, we have a free variable in the system. Let \(x_3 = t\) where \(t\) is a parameter. Solve for \(x_1\) and \(x_2\) in terms of \(t\).From row 2: \(0 + x_2 + \frac{3}{2}t = 0 \rightarrow x_2 = -\frac{3}{2}t\).Substitute into row 1: \(x_1 - (-\frac{3}{2}t) - 2t = 0 \rightarrow x_1 = \frac{7}{2}t\).Thus, the solution is:\[x_1 = \frac{7}{2}t, \quad x_2 = -\frac{3}{2}t, \quad x_3 = t.\]

Step 6: Interpret the Results

This system, therefore, has infinitely many solutions, each defined by the parameter \(t\). The solutions form a line in three-dimensional space parameterized by \(t\). Each value of \(t\) gives a different solution to the system.

Key Concepts

Gauss-Jordan elimination

Gauss-Jordan elimination is a method used in linear algebra to solve systems of linear equations. It enhances the process of Gaussian elimination to reach a reduced row echelon form (RREF).
The idea is to transform the augmented matrix of the system into an identity matrix—or as close as it can be—with corresponding solutions. This makes finding the solutions straightforward.

• First, you'll want to create zeros below (and above, in the case of Gauss-Jordan elimination) each pivot in the matrix.
• Next, each row should be simplified by making the leading coefficient (pivot) equal to one.
• This can involve row swapping, scaling rows, or adding/subtracting multiples of rows.

When each column of a variable has one non-zero number, it will often be either 1 or 0. Using Gauss-Jordan elimination simplifies discovering solutions to systems with a clearer picture of the relationship between variables.

augmented matrix

An augmented matrix is a compact way to represent a system of linear equations. It's created by appending the constant terms of the equations to the right side of the coefficient matrix, using a vertical line to separate them.
For example, in the given system, we have:

• \(x_1 - x_2 - 2x_3 = 0\)
• \(2x_1 + 4x_2 + 5x_3 = 0\)
• \(6x_1 - 3x_3 = 0\)

The augmented matrix is:
\[\begin{bmatrix} 1 & -1 & -2 & | & 0 \ 2 & 4 & 5 & | & 0 \ 6 & 0 & -3 & | & 0 \end{bmatrix} \]
This matrix can then be manipulated using row operations as part of Gaussian or Gauss-Jordan elimination methods. This approach allows for a structured path to systematically determine the solutions or show if none exist.

infinitely many solutions

In the context of linear algebra, a system having infinitely many solutions often implies that there's more than one way to satisfy all equations simultaneously.
In our example, through reduction of the matrix, we find a row of zeros, such as: \[\begin{bmatrix} 0 & 0 & 0 & | & 0 \end{bmatrix}\]
This indicates a dependence amongst the variables—meaning one or more variables are free.
Free variables can take any value, allowing corresponding adjustments in other variables to satisfy the equations. As demonstrated, in our simplified equation form:

• Let \(x_3 = t\), a parameter representing infinitely many choices.
• Then, solve for other variables in terms of "t", such as \(x_1 = \frac{7}{2}t\), and \(x_2 = -\frac{3}{2}t\).

Thus, each value of \(t\) in these equations represents a specific line in three-dimensional space, showcasing the concept of having infinitely many solutions.

linear algebra

Linear algebra is the branch of mathematics that deals predominantly with vectors, vector spaces, and linear equations. It provides tools to solve systems of linear equations efficiently and forms the foundation for much of modern mathematics and engineering applications.

• It involves concepts like matrices, determinants, vector spaces, and eigenvalues.
• Techniques including Gaussian and Gauss-Jordan elimination make solving large systems simpler.
• The subject helps to understand the structure and properties of vector spaces and transformations represented by matrices.

These concepts form the backbone of fields such as computer graphics, machine learning, and operations research. Solving systems using linear algebra not only finds numerical solutions but also facilitates understanding the geometric and theoretical aspects of the relationships among variables in linear equations.