Problem 25 - Algebra and Trigonometry

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x_{1}+2 x_{2}-x_{3} &=9 \\ 2 x_{1} -x_{3}=-2 \\ 3 x_{1}+5 x_{2}+2 x_{3} &=22 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The unique solution is \( x_1 = -1, x_2 = 5, x_3 = 0 \).

Step 1: Write the Augmented Matrix

We start by writing the given system of equations as an augmented matrix. The system is: \(x_1 + 2x_2 - x_3 = 9\), \(2x_1 - x_3 = -2\), and \(3x_1 + 5x_2 + 2x_3 = 22\). The corresponding augmented matrix is: \[\begin{bmatrix}1 & 2 & -1 & | & 9 \2 & 0 & -1 & | & -2 \3 & 5 & 2 & | & 22 \\end{bmatrix}\]

Step 2: Apply Row Operations to Form Upper Triangular Matrix

The first step in the Gauss-Jordan elimination technique is to transform the matrix into an upper triangular form. We use row operations to eliminate variables step by step. - Replace Row 2 (R2) with \(R2 - 2 \times R1\): \(\begin{bmatrix} 2 & 0 & -1 & | & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 2 & -1 & | & 9 \end{bmatrix}\) leading to \(\begin{bmatrix} 0 & -4 & 1 & | & -20 \end{bmatrix}\) - Replace Row 3 (R3) with \(R3 - 3 \times R1\): \(\begin{bmatrix} 3 & 5 & 2 & | & 22 \end{bmatrix} - 3 \begin{bmatrix} 1 & 2 & -1 & | & 9 \end{bmatrix}\) leading to \(\begin{bmatrix} 0 & -1 & 5 & | & -5 \end{bmatrix}\)The matrix now looks like:\[\begin{bmatrix}1 & 2 & -1 & | & 9 \0 & -4 & 1 & | & -20 \0 & -1 & 5 & | & -5 \\end{bmatrix}\]

Step 3: Continue Applying Row Operations

Further simplify the matrix to reach reduced row-echelon form:- Replace Row 3 (R3) with \(R3 - \frac{1}{4} \times R2\) which eliminates the coefficient of \(x_2\) in R3: \(\begin{bmatrix} 0 & -1 & 5 & | & -5 \end{bmatrix} + \frac{1}{4} \begin{bmatrix} 0 & -4 & 1 & | & -20 \end{bmatrix}\) becomes \(\begin{bmatrix} 0 & 0 & 9/4 & | & 0 \end{bmatrix}\)Update the matrix:\[\begin{bmatrix}1 & 2 & -1 & | & 9 \0 & -4 & 1 & | & -20 \0 & 0 & 9/4 & | & 0 \\end{bmatrix}\]

Step 4: Back-Substitution

From the reduced form of the matrix, back-substitute to find each variable: - Solve for \(x_3\): \[ \frac{9}{4} x_3 = 0 \implies x_3 = 0 \]- Substitute \(x_3 = 0\) into the equation from R2: \[ -4x_2 + x_3 = -20 \implies -4x_2 = -20 \implies x_2 = 5 \]- Substitute \(x_2 = 5\) and \(x_3 = 0\) into the equation from R1: \[ x_1 + 2(5) - 0 = 9 \implies x_1 + 10 = 9 \implies x_1 = -1 \]

Step 5: State the Solution

Now that we have solved each variable, we can state the solution of the system as:\( x_1 = -1, x_2 = 5, x_3 = 0 \). The solution is unique, confirming that the system is consistent and independent.

Key Concepts

Gauss-Jordan elimination

Gauss-Jordan elimination is a method for solving systems of linear equations. It involves performing operations on an augmented matrix until it is in reduced row-echelon form. The main purpose of this elimination method is to simplify the system of equations, making it much easier to solve. By transforming the matrix using this method, each leading entry (or pivot) in the rows becomes one, and all other entries in the column of a pivot are zeros. This transformation essentially aims to clear each column under the pivots, ensuring that the system can be easily interpreted for solutions.

• The initial goal is to create an upper triangular form and then further reduce it to the identity matrix
• At each step, row operations such as adding, subtracting, and multiplying rows are used to maneuver the matrix into the desired shape

This is the systematic difference between Gaussian elimination and Gauss-Jordan elimination, the latter requiring full reduction to the identity matrix for straightforward reading of the solutions.

augmented matrix

An augmented matrix is a convenient way to represent a system of linear equations. It combines the coefficients of the variables and the constants from the equations into a single matrix format.
This matrix includes the variables' coefficients on the left and the constants on the right side of a delimiter like the vertical line (|).
For instance, consider the following system of equations: \(x_1 + 2x_2 - x_3 = 9\), \(2x_1 - x_3 = -2\), and \(3x_1 + 5x_2 + 2x_3 = 22\). The corresponding augmented matrix would be: \[\begin{bmatrix} 1 & 2 & -1 & | & 9 \ 2 & 0 & -1 & | & -2 \ 3 & 5 & 2 & | & 22 \end{bmatrix}\]

• The purpose of the augmented matrix is to simplify the representation and manipulation of the system
• It aligns all the equations, so that operations on rows can be easily tracked

This format lays the groundwork for applying the Gauss-Jordan elimination technique.

row operations

Row operations are fundamental in transforming an augmented matrix into its desired form. Row operations consist of three main types that can be applied to any matrix:

• Switching two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting the multiple of one row to another row

These operations are designed to create zeros below each pivot (the first non-zero entry in a row from the first column to the last). That means clearing out entire columns to simplify and solve the system.
For instance, when transforming the matrix in our problem,

• Row 2 (R2) was replaced with \(R2 - 2 \times R1\) to create a zero in the first column
• Similarly, Row 3 (R3) was adjusted to help move towards an upper triangular matrix

These steps are repeated until the matrix reaches reduced row-echelon form.

back-substitution

Once the matrix is in reduced row-echelon form, back-substitution is used to find the values of the variables. This process involves solving the equations from the bottom row up to the top row, from the last variable back to the first. In our system, after simplifying down to:\[\begin{bmatrix} 1 & 2 & -1 & | & 9 \ 0 & -4 & 1 & | & -20 \ 0 & 0 & \frac{9}{4} & | & 0 \end{bmatrix}\]

• First, solve for \(x_3\) directly from the third equation: \(\frac{9}{4} x_3 = 0\), which means \(x_3 = 0\)
• Then, substitute \(x_3\) into the second equation to find \(x_2\)
• Use \(x_2\) and \(x_3\) values in the first equation to find \(x_1\)

By systematically solving from the bottom row up, each step becomes clear, and the unique solution reveals itself: \(x_1 = -1\), \(x_2 = 5\), and \(x_3 = 0\).