Problem 65 - College Algebra with Modeling and Visualization

Question

Reduced Row-Rchelon Form Use Gauss-Jordan elimination to solve the system of equations. $$ \begin{aligned} 2 x+y-z &=2 \\ x-2 y+z &=0 \\ x+3 y-2 z &=4 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{7}{3}\), \(y = \frac{10}{3}\), \(z = 4\).

Step 1: Write the Augmented Matrix

Start by writing the system of equations as an augmented matrix. For the given system, the augmented matrix is:\[\begin{bmatrix}2 & 1 & -1 & | & 2 \1 & -2 & 1 & | & 0 \1 & 3 & -2 & | & 4 \\end{bmatrix}\]Each row corresponds to one equation, with the coefficients and constants arranged accordingly.

Step 2: Perform Row Operations (Form Triangular Matrix)

We aim to form an upper triangular matrix. Start by subtracting Row 2 from Row 1 to make the first element in Row 2 zero:\[R2 \rightarrow R2 - R1 \ \begin{bmatrix}2 & 1 & -1 & | & 2 \0 & -3 & 2 & | & -2 \1 & 3 & -2 & | & 4 \\end{bmatrix}\]Next, subtract Row 3 from Row 1 to make the first element in Row 3 zero:\[R3 \rightarrow R3 - \frac{1}{2}R1 \ \begin{bmatrix}2 & 1 & -1 & | & 2 \0 & -3 & 2 & | & -2 \0 & 2.5 & -1.5 & | & 3 \\end{bmatrix}\]

Step 3: Make the Lead Coefficient of Row 2 to 1

To simplify, make row two's first non-zero element (the leading coefficient) equal to one by dividing the whole row by -3:\[R2 \rightarrow \frac{1}{-3}R2 \ \begin{bmatrix}2 & 1 & -1 & | & 2 \0 & 1 & -\frac{2}{3} & | & \frac{2}{3} \0 & 2.5 & -1.5 & | & 3 \\end{bmatrix}\]

Step 4: Eliminate Coefficients below Leading 1 of Row 2

Eliminate the coefficients below the leading 1 in Row 2 by performing the operation on Row 3:\[R3 \rightarrow R3 - 2.5R2 \ \begin{bmatrix}2 & 1 & -1 & | & 2 \0 & 1 & -\frac{2}{3} & | & \frac{2}{3} \0 & 0 & 0.833 & | & 3.333 \\end{bmatrix}\]

Step 5: Normalize Row 3's Leading Coefficient

Divide Row 3 by 0.833 to make the leading coefficient 1:\[R3 \rightarrow \frac{1}{0.833}R3 \ \begin{bmatrix}2 & 1 & -1 & | & 2 \0 & 1 & -\frac{2}{3} & | & \frac{2}{3} \0 & 0 & 1 & | & 4 \\end{bmatrix}\]

Step 6: Back Substitution to Row 2 and Row 1

Eliminate non-zero values above the leading 1 of Row 3 (for Row 1 and Row 2), starting with Row 2:\[R2 \rightarrow R2 + \frac{2}{3}R3 \\begin{bmatrix}2 & 1 & -1 & | & 2 \0 & 1 & 0 & | & \frac{10}{3} \0 & 0 & 1 & | & 4 \\end{bmatrix}\]Then for Row 1:\[R1 \rightarrow R1 + R3 \\begin{bmatrix}2 & 1 & 0 & | & 6 \0 & 1 & 0 & | & \frac{10}{3} \0 & 0 & 1 & | & 4 \\end{bmatrix}\]Next, eliminate the x-term in Row 1:\[R1 \rightarrow \frac{1}{2}R1 \\begin{bmatrix}1 & 0.5 & 0 & | & 3 \0 & 1 & 0 & | & \frac{10}{3} \0 & 0 & 1 & | & 4 \\end{bmatrix}\]Further eliminate the y-coefficient in Row 1:\[R1 \rightarrow R1 - 0.5R2 \\begin{bmatrix}1 & 0 & 0 & | & \frac{7}{3} \0 & 1 & 0 & | & \frac{10}{3} \0 & 0 & 1 & | & 4 \\end{bmatrix}\]

Step 7: Interpret the Solution

From the final reduced row-echelon form matrix, we can read off the solution for \(x\), \(y\), and \(z\):\[x = \frac{7}{3}, \quad y = \frac{10}{3}, \quad z = 4\]Thus, the solution to the system of equations is \(x = \frac{7}{3}\), \(y = \frac{10}{3}\), and \(z = 4\).

Key Concepts

Augmented Matrix

In linear algebra, the augmented matrix is a powerful tool to represent a system of linear equations. By converting the equations into a matrix form, it simplifies the process of solving the system, especially using techniques like Gauss-Jordan elimination. Each row in the augmented matrix corresponds to an equation in the system.
For example, consider the system of equations:

• \( 2x + y - z = 2 \)
• \( x - 2y + z = 0 \)
• \( x + 3y - 2z = 4 \)

When we write this system as an augmented matrix, it looks like this:\[\begin{bmatrix}2 & 1 & -1 & | & 2 \1 & -2 & 1 & | & 0 \1 & 3 & -2 & | & 4 \\end{bmatrix}\] The numbers before the vertical line represent the coefficients of the variables in each equation, and the numbers after the line represent the constants. This matrix makes it easier to apply row operations to eventually solve for the variables.

Reduced Row-Echelon Form

The Reduced Row-Echelon Form (RREF) is a special type of matrix form used to solve systems of linear equations. Once a matrix is in RREF, you can easily identify the solution to the system. A matrix is said to be in reduced row-echelon form if:

• The leading entry (first non-zero number from the left) in each row is 1.
• Each leading 1 is the only non-zero entry in its column.
• The leading 1 in any row is to the right of any leading 1s in previous rows.
• Any rows containing only zeros are at the bottom of the matrix.

Using Gauss-Jordan elimination, you transform your matrix step by step into this form. From the example of the system given, the final reduced row-echelon form is: \[\begin{bmatrix}1 & 0 & 0 & | & \frac{7}{3} \0 & 1 & 0 & | & \frac{10}{3} \0 & 0 & 1 & | & 4 \\end{bmatrix}\] This form allows us to directly conclude the values of the variables: \( x = \frac{7}{3} \), \( y = \frac{10}{3} \), and \( z = 4 \).

System of Equations

A system of equations is a set of two or more equations involving the same set of variables. Solving such a system means finding values for the variables that satisfy all equations simultaneously. For example, in our given problem, the system is:

• \( 2x + y - z = 2 \)
• \( x - 2y + z = 0 \)
• \( x + 3y - 2z = 4 \)

To solve this system, we can use the method of Gauss-Jordan elimination, where we systematically eliminate variables by performing row operations to transform the system's augmented matrix into reduced row-echelon form. This process ensures the equations simplify from a system of three equations to individual, isolated equations for each variable, thus making it straightforward to determine the solution.

Linear Algebra

Linear algebra is a branch of mathematics focused on vector spaces and the linear mappings between these spaces. It is foundational for understanding and solving linear systems and analyzing mathematical properties using matrices and vectors.
In the context of solving systems of equations, linear algebra provides efficient methods like Gaussian elimination and Gauss-Jordan elimination to simplify complex systems. By using matrix representations for systems of equations, we can apply these algorithms to find solutions much faster than traditional methods.
These techniques are invaluable in various applications across fields such as computer science, engineering, and economics, where systems of linear equations frequently arise. Understanding concepts such as the augmented matrix and reduced row-echelon form within linear algebra enables students to handle a wide range of problems involving linear relationships.