Problem 1 - Advanced Engineering Mathematics

Question

Use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. \(\begin{aligned} x_{1}-x_{2} &=11 \\ 4 x_{1}+3 x_{2} &=-5 \end{aligned}\)

Step-by-Step Solution

Answer

The solution is \(x_1 = 4\) and \(x_2 = -7\).

Step 1: Write the Augmented Matrix

First, we write the system of equations as an augmented matrix. The given system is: \( \begin{aligned} x_{1}-x_{2} &=11 \ 4x_{1}+3x_{2} &=-5 \end{aligned} \). This can be written as the augmented matrix: \[ \begin{bmatrix} 1 & -1 & | & 11 \ 4 & 3 & | & -5 \end{bmatrix} \]

Step 2: Perform Row Operations to Achieve a Row-Echelon Form

We need to use row operations to create zeros below the first entry in the first column. We notice Row 1 has a 1, which can stay unchanged. Multiply Row 1 by 4 and subtract Row 2: \[ \begin{bmatrix} 1 & -1 & | & 11 \ 0 & 7 & | & -49 \end{bmatrix} \]. Now, the matrix is in row-echelon form.

Step 3: Solve for Variables from Row-Echelon Form

From the second row \(0x_1 + 7x_2 = -49\), we simplify to find \(x_2\). \(7x_2 = -49 \Rightarrow x_2 = -7\). Substitute \(x_2 = -7\) into the first equation \(x_1 - x_2 = 11\): \(x_1 - (-7) = 11 \Rightarrow x_1 = 4\).

Step 4: Verify the Solution

Verify the values: \(x_1 = 4\) and \(x_2 = -7\) by substituting back into the original equations: 1. \(4 - (-7) = 11\), which is correct. 2. \(4(4) + 3(-7) = -5\), which also holds true. The values verify both equations, confirming the solution.

Key Concepts

Gauss-Jordan elimination

Gauss-Jordan elimination is an algorithm used in linear algebra to solve systems of linear equations. It's an extension of Gaussian elimination, with the added step of creating zeros both below and above the pivot positions in each column. This results in a matrix called its reduced row-echelon form, where each leading entry is 1 and is the only non-zero entry in its column.

Gauss-Jordan elimination offers a systematic way to find the precise solution to a system of equations or to verify if a unique or no solution exists. Compared to Gaussian elimination, Gauss-Jordan elimination simplifies all steps by continuing the row operations until the matrix is in its simplest form.

It's important to:

• Choose pivot columns strategically.
• Conduct row operations carefully to avoid errors.

This technique can efficiently handle complex systems of equations by providing clear, step-by-step transformations of the matrix.

augmented matrix

An augmented matrix is a compact way of representing a system of linear equations. It combines the coefficients of the variables and the constants from the equations into a single matrix format.

This kind of matrix separates the variables and their respective coefficients from the constants using a vertical line, known as the augmentation bar. For example, the system given in the problem can be represented as an augmented matrix:

• The first row represents the equation \(x_1 - x_2 = 11\), which translates into matrix form as \([1, -1 | 11]\).
• The second row, \(4x_1 + 3x_2 = -5\), becomes \([4, 3 | -5]\).

Using this matrix transformation simplifies the manipulations required during the elimination process, allowing us to focus on the row operations needed to solve the system.

row-echelon form

The row-echelon form (REF) of a matrix is a key step in solving systems of equations through elimination methods. A matrix is in REF when:

• All non-zero rows are above any rows full of zeros.
• The leading entry (also called a pivot) of each non-zero row is to the right of the leading entry in the previous row.

This means each pivot creates a staircase-like pattern across the matrix. In our problem, once the row operations are performed, we obtain the following REF:\[\begin{bmatrix}1 & -1 & | & 11 \0 & 7 & | & -49\end{bmatrix}\]The main advantage of converting to REF is to simplify the equations systematically, allowing easier back-substitution to find solutions.

solving systems of equations

Solving systems of equations involves finding values for the variables that satisfy all equations simultaneously. This can be done using methods like substitution, elimination, or matrix operations.

In the case of the problem we tackled, we performed Gaussian elimination, progressing to row-echelon form, and then solved the system through back-substitution:

• From equation \(0x_1 + 7x_2 = -49\), it was straightforward to apply basic algebra and arrive at \(x_2 = -7\).
• Substituting \(x_2 = -7\) into the first equation, we found \(x_1 = 4\).

To ensure the solution's accuracy, verifying each value by substituting back into the original equations is necessary.

Solutions indicate the intersection point of the equations or confirm if the lines are parallel, which typically imply no solution exists.