Problem 12 - Advanced Engineering Mathematics

Question

In Problems 1-20, use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. $$ \begin{array}{r} x_{1}-x_{2}-2 x_{3}=0 \\ 2 x_{1}+4 x_{2}+5 x_{3}=0 \\ 6 x_{1}-\quad 3 x_{3}=0 \end{array} $$

Step-by-Step Solution

Answer

The system has infinitely many solutions; parametrize with \(x_1 = -\frac{1}{2}t, x_2 = -\frac{3}{2}t, x_3 = t\).

Step 1: Write the Augmented Matrix

Begin by writing the given system of equations as an augmented matrix. The system is:\[\begin{align*}x_{1} - x_{2} - 2x_{3} &= 0 \2x_{1} + 4x_{2} + 5x_{3} &= 0 \6x_{1} - 3x_{3} &= 0\end{align*}\]This corresponds to the augmented matrix:\[\begin{bmatrix}1 & -1 & -2 & | & 0 \2 & 4 & 5 & | & 0 \6 & 0 & -3 & | & 0\end{bmatrix}\]

Step 2: Row Reduce to Row Echelon Form (REF)

Use Gaussian elimination to convert the matrix to row echelon form.1. Eliminate the first coefficient under pivot in the first column by replacing Row 2 with \( -2 \times \text{Row 1} + \text{Row 2} \):\[\begin{bmatrix}1 & -1 & -2 & | & 0 \0 & 6 & 9 & | & 0 \6 & 0 & -3 & | & 0\end{bmatrix}\]2. Eliminate the second coefficient under pivot in the first column by replacing Row 3 with \( -6 \times \text{Row 1} + \text{Row 3} \):\[\begin{bmatrix}1 & -1 & -2 & | & 0 \0 & 6 & 9 & | & 0 \0 & 6 & 9 & | & 0\end{bmatrix}\]

Step 3: Simplify the Row Echelon Form

Simplify Row 3 by replacing it with \( \text{Row 3} - \text{Row 2} \):\[\begin{bmatrix}1 & -1 & -2 & | & 0 \0 & 6 & 9 & | & 0 \0 & 0 & 0 & | & 0\end{bmatrix}\]

Step 4: Convert to Reduced Row Echelon Form (RREF)

Convert the matrix to reduced row echelon form.1. Make the leading coefficient of Row 2 a 1 by dividing the entire row by 6:\[\begin{bmatrix}1 & -1 & -2 & | & 0 \0 & 1 & \frac{3}{2} & | & 0 \0 & 0 & 0 & | & 0\end{bmatrix}\]2. Eliminate the coefficient in Row 1, Column 2 by replacing Row 1 with \( \text{Row 1} + \text{Row 2} \):\[\begin{bmatrix}1 & 0 & \frac{1}{2} & | & 0 \0 & 1 & \frac{3}{2} & | & 0 \0 & 0 & 0 & | & 0\end{bmatrix}\]

Step 5: Determine the Solution

The system in reduced row echelon form indicates one free variable:- \(x_3 = t\) (where \(t\) is a parameter representing any real number)- From Row 2: \(x_2 + \frac{3}{2}x_3 = 0\) implies \(x_2 = -\frac{3}{2}t\)- From Row 1: \(x_1 + \frac{1}{2}x_3 = 0\) implies \(x_1 = -\frac{1}{2}t\)The solution is given parametrically by:\[x_1 = -\frac{1}{2}t, \quad x_2 = -\frac{3}{2}t, \quad x_3 = t\]

Key Concepts

Gauss-Jordan Elimination

Gauss-Jordan elimination is a method used to solve systems of linear equations. It goes a step further than Gaussian elimination by transforming a matrix into reduced row echelon form.

The process involves both row operations and strategic manipulations to achieve a simpler matrix form. Here’s how it works:

• Row Operations: You can swap rows, multiply a row by a non-zero constant, or add/subtract multiples of rows to each other. These changes systematically simplify the matrix.
• Stopping Point: Gauss-Jordan elimination continues until you reach a matrix where each leading coefficient (also known as a pivot) is 1, and all other elements in the pivot’s column are 0.

The advantage of using Gauss-Jordan elimination is that it gives a more straightforward form of the system called the Reduced Row Echelon Form (RREF), which is extremely useful for finding solutions to equations.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. Instead of writing out all the equations separately, you can combine the coefficients and constants into one single matrix form.

This representation is organized by placing the coefficients of the variables into columns and appending an extra column for the constants. So, for a system like:

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

you create an augmented matrix as: \[\begin{bmatrix}1 & -1 & -2 & | & 0 \2 & 4 & 5 & | & 0 \6 & 0 & -3 & | & 0\end{bmatrix}\]With all the equations embedded into this matrix, it's much easier to perform operations like Gaussian and Gauss-Jordan eliminations on it.

Row Echelon Form

Row echelon form (REF) is an intermediate step in matrix transformations, used to make solving systems of equations easier. In REF:

• Each row starts with zeros until a non-zero pivot number appears.
• The pivot number leads the row.
• The number of leading zeros increases down the rows.

Reaching the row echelon form helps simplify the process of solving equations by transforming the original system into a form that's more straightforward to interpret and solve with back-substitution.

During this process, you manipulate the matrix using row operations to create zeros below each leading pivot, turning the matrix into a triangular shape. For instance, after applying Gaussian elimination, our example matrix becomes:\[\begin{bmatrix}1 & -1 & -2 & | & 0 \0 & 6 & 9 & | & 0 \0 & 0 & 0 & | & 0\end{bmatrix}\] This form reduces the complexity of solving the system and provides a solid foundation for finding the solution.

Reduced Row Echelon Form

The reduced row echelon form (RREF) is the final, simplified matrix form achieved by applying Gauss-Jordan elimination. Achieving RREF makes it straightforward to identify solutions to the system.

Characterized by:

• The leading entry in each non-zero row is 1, also called a pivot.
• Each pivot entry is the only non-zero number in its column, creating zeros above and below the pivot.

For example, transforming the row echelon form into RREF for our system, the matrix becomes:\[\begin{bmatrix}1 & 0 & \frac{1}{2} & | & 0 \0 & 1 & \frac{3}{2} & | & 0 \0 & 0 & 0 & | & 0\end{bmatrix}\]This powerful form reveals immediate solutions or connections between variables. It indicates directly how some variables depend on others or shows free variables, like in our case with \(x_3 = t\). Through RREF, understanding and solving systems of equations becomes clearer and more straightforward.