Problem 29 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the system of linear equations, using the Gauss-Jordan elimination method. \(2 x+y-3 z=1\) \(x-y+2 z=1\) \(5 x-2 y+3 z=6\)

Step-by-Step Solution

Answer

The solution to the system of linear equations using the Gauss-Jordan elimination method is (x, y, z) = (0.75, 1, 3/4).

Step 1: Write the augmented matrix for the system of equations

To form the augmented matrix, we will write the coefficients of x, y, and z as well as the constants in an [A|B] format. The given system of equations can be represented as follows: \[ \begin{bmatrix} 2 & 1 & -3 & | & 1 \\ 1 & -1 & 2 & | & 1 \\ 5 & -2 & 3 & | & 6 \end{bmatrix} \]

Step 2: Obtain a zero below the pivot in the first column

We will make the element in row 2 and column 1 zero by replacing the second row with the subtraction of the second row from half of the first row. Here, the pivot (the first non-zero entry in the first row) is 2. \[ \begin{bmatrix} 2 & 1 & -3 & | & 1 \\ 0 & -\frac{3}{2} & \frac{7}{2} & | & \frac{1}{2} \\ 5 & -2 & 3 & | & 6 \end{bmatrix} \]

Step 3: Get a zero below the pivot in the first column

Now, we will make the element in row 3 and column 1 zero by replacing the third row with the subtraction of 2.5 times the first row from the third row. \[ \begin{bmatrix} 2 & 1 & -3 & | & 1 \\ 0 & -\frac{3}{2} & \frac{7}{2} & | & \frac{1}{2} \\ 0 & -\frac{5}{2} & \frac{21}{2} & | & \frac{11}{2} \end{bmatrix} \]

Step 4: Get a zero below the pivot in the second column

Now, we will make the element in row 3 and column 2 zero by replacing the third row with the subtraction of 1.6667 times the second row from the third row. \[ \begin{bmatrix} 2 & 1 & -3 & | & 1 \\ 0 & -\frac{3}{2} & \frac{7}{2} & | & \frac{1}{2} \\ 0 & 0 & 4 & | & 3 \end{bmatrix} \]

Step 5: Convert the matrix to Reduced Row Echelon Form (RREF)

We will transform the matrix into RREF by scaling each row so that the leading entry of each row, called the pivot, is 1. \[ \begin{bmatrix} 1 & 0.5 & -1.5 & | & 0.5 \\ 0 & 1 & -\frac{7}{3} & | & -\frac{1}{3} \\ 0 & 0 & 1 & | & \frac{3}{4} \end{bmatrix} \] Now, get zeros above the pivots in columns 2 and 3: \[ \begin{bmatrix} 1 & 0.5 & 0 & | & 1.25 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & \frac{3}{4} \end{bmatrix} \] Finally, get a zero above the pivot in column 1: \[ \begin{bmatrix} 1 & 0 & 0 & | & 0.75 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & \frac{3}{4} \end{bmatrix} \]

Step 6: Write the solution as an ordered triple

From the RREF, we now read the solution as an ordered triple (x, y, z): (0.75, 1, 3/4)

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple equations with several variables. The goal is to find a set of values for the variables that satisfy all the equations simultaneously. Consider the following system:

• \( 2x + y - 3z = 1 \)
• \( x - y + 2z = 1 \)
• \( 5x - 2y + 3z = 6 \)

In this system, we have three equations involving three variables \( x, y, \) and \( z \). Solving the system means finding the specific values of \( x, y, \) and \( z \) that make all three equations true. Systems of linear equations are often solved using systematic methods like substitution, elimination, or matrix techniques, such as the Gauss-Jordan elimination method. This process allows us to simplify the system and find an exact or approximate solution.

Augmented Matrix

The augmented matrix is a matrix representation used to solve a system of linear equations. It combines the coefficients of the variables and the constants from each equation into a single matrix. This makes it easier to manipulate equations mathematically. For example, in the given system:

• Coefficients of \( x \) are found in the first column.
• Coefficients of \( y \) are in the second column.
• Coefficients of \( z \) are in the third column.
• The constants are in the last column, separated by a vertical bar \(|\) from the coefficients.

The augmented matrix for our equations looks like this:\[\begin{bmatrix} 2 & 1 & -3 & | & 1 \ 1 & -1 & 2 & | & 1 \ 5 & -2 & 3 & | & 6\end{bmatrix}\]This format is compact and efficient for applying operations like row operations, which are crucial for techniques like the Gauss-Jordan elimination. By working with augmented matrices, we can more easily perform systematic manipulation needed to find the solution to the system.

Reduced Row Echelon Form (RREF)

Reduced Row Echelon Form (RREF) is a specific type of matrix form that facilitates solving linear equations. A matrix is in RREF when:

• Each leading entry (the first non-zero number from the left) in a row is 1.
• Each leading 1 is the only non-zero entry in its column.
• The leading 1 in a lower row is to the right of the leading 1 of the row above it.
• Any rows of all zeros are at the bottom of the matrix.

Applying the Gauss-Jordan elimination method transforms an augmented matrix into its RREF. This is done by:- Making pivots (leading entries) 1 by scaling rows.- Eliminating numbers above and below pivots to make them zeros.Our given system's RREF is:\[\begin{bmatrix} 1 & 0 & 0 & | & 0.75 \ 0 & 1 & 0 & | & 1 \ 0 & 0 & 1 & | & \frac{3}{4}\end{bmatrix}\]From this form, we directly obtain the solution to the system: \( x = 0.75, y = 1, z = \frac{3}{4} \). RREF allows us to quickly and precisely read the values of the unknowns, making it a powerful tool in linear algebra.