Problem 39 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the system of linear equations using the Gauss-Jordan elimination method. \(\begin{aligned} x+y+z &=0 \\ 2 x-y+z &=1 \\ x+y-2 z &=2 \end{aligned}\)

Step-by-Step Solution

Answer

In short, given the system of linear equations: \(\begin{aligned} x + y + z = 0 \\ 2x - y + z = 1 \\ x + y - 2z = 2 \end{aligned}\) We can solve it using the Gauss-Jordan elimination method, by performing row operations to reach reduced row echelon form (RREF). The final RREF augmented matrix is: \( \left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array}\right] \) The solution for this system is \(x = 1\), \(y = -1\), and \(z = 0\).

Step 1: Write the augmented matrix

To represent the given system of linear equations as an augmented matrix, we will list the coefficients of x, y, and z from each equation, followed by the constant term. \( \left[\begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 2 & -1 & 1 & 1 \\ 1 & 1 & -2 & 2 \end{array}\right] \)

Step 2: Perform row operations to reach RREF

We will now perform the following row operations: 1. Add -2 times Row 1 to Row 2 to eliminate x from the second equation. 2. Add -1 times Row 1 to Row 3 to eliminate x from the third equation. 3. Divide Row 2 by 3 to get a leading 1 in the second row. 4. Add Row 2 to Row 1 to eliminate y from the first equation. 5. Add -2 times Row 2 to Row 3 to eliminate y from the third equation. 6. Divide Row 3 by 3 to get a leading 1 in the third row. 7. Add -1 times Row 3 to Row 1 to eliminate z from the first equation. 8. Add Row 3 to Row 2 to eliminate z from the second equation. \( \left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array}\right] \)

Step 3: Write the solutions for x, y, and z

Now that we have the augmented matrix in RREF, we can convert it back to a system of linear equations and write the solution: \(x = 1, y=-1, z=0\)

Key Concepts

Linear Equations

Linear equations are mathematical expressions that imply relationships of equilibrium. These are equations where each term is either a constant or the product of a constant and a single variable. In the realm of algebra, solving linear equations is akin to finding points of intersection, be it on a line or within a multidimensional space.
A system of linear equations might consist of multiple such equations that we try to solve simultaneously. For example, the system given by the exercise can be represented as:

• \(x + y + z = 0\)
• \(2x - y + z = 1\)
• \(x + y - 2z = 2\)

Each equation in the system represents a plane in three-dimensional space, and the solution to the system is the point or points where these planes intersect.

Augmented Matrix

To simplify the process of solving a system of linear equations, we use an augmented matrix. This involves restructuring our equations into a matrix form. Specifically, it captures the coefficients of each variable and the constants on the right side of the equation.
In the given example, the augmented matrix is represented as:
\[\begin{bmatrix} 1 & 1 & 1 & \vert & 0 \2 & -1 & 1 & \vert & 1 \1 & 1 & -2 & \vert & 2 \end{bmatrix}\]
The vertical line separates the coefficients of the variables from the constants; this layout helps organize the data efficiently and prepares it for row operations.

Row Operations

Row operations are crucial techniques used to manipulate matrices into simpler forms, such as the Row-Reduced Echelon Form (RREF). There are three main types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero scalar.
• Adding or subtracting the multiple of one row from another.

These operations are elementary yet powerful tools that maintain the equivalence of the original system. Applying them strategically will lead us to simplify the matrix until it reaches the RREF, making it easy to determine the solutions.
In the example, a series of strategic operations were applied to transform the initial augmented matrix step by step, refining it to its simplest form.

RREF

Reaching the Row-Reduced Echelon Form (RREF) of a matrix is a key goal when solving a system of equations. An RREF matrix has specific characteristics:

• Leading entries in each row are 1, also known as pivots.
• Each leading 1 is the only non-zero entry in its column.
• Zeros are above and below each leading 1.
• Each subsequent row has its leading 1 further to the right.

The transformation of the augmented matrix to RREF effectively provides a straightforward way to read off the solutions to the system. In our example, transforming the matrix using row operations resulted in:
\[\begin{bmatrix} 1 & 0 & 0 & \vert & 1 \0 & 1 & 0 & \vert & -1 \0 & 0 & 1 & \vert & 0 \end{bmatrix}\]

Solution of System of Equations

Once we have our matrix in Row-Reduced Echelon Form, it is time to interpret the matrix back into linear equations. This step helps us determine the values of the variables in the original system of equations.
The RREF of our matrix is:
\[\begin{bmatrix} 1 & 0 & 0 & \vert & 1 \0 & 1 & 0 & \vert & -1 \0 & 0 & 1 & \vert & 0 \end{bmatrix}\]
Translating back to equations, this matrix reveals:

• \(x = 1\)
• \(y = -1\)
• \(z = 0\)

These are the unique solutions for the variables in the system. We concluded that each variable has a precise value that satisfies all the original linear equations simultaneously.