Problem 39 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l} 1.1 x+0.7 y-3.1 z=-1.79 \\ 2.1 x+0.5 y-1.6 z=-0.13 \\ 0.5 x+0.4 y-0.5 z=-0.07 \end{array} $$

Step-by-Step Solution

Answer

The solution is approximately \(x = 1.4\), \(y = -0.569\), \(z = 0.8\).

Step 1: Write the system in augmented matrix form

Convert the system of equations into an augmented matrix. This involves writing the coefficients of the variables and the constants into a matrix: \[\begin{bmatrix}1.1 & 0.7 & -3.1 & | & -1.79 \2.1 & 0.5 & -1.6 & | & -0.13 \0.5 & 0.4 & -0.5 & | & -0.07\end{bmatrix}\]

Step 2: Use row operations to form a triangular matrix

Perform row operations to make the elements below the first pivot (1.1) zero. 1. Subtract \(2.1/1.1\) times the first row from the second row.2. Subtract \(0.5/1.1\) times the first row from the third row.This yields:\[\begin{bmatrix}1.1 & 0.7 & -3.1 & | & -1.79 \0 & -0.832 & 3.523 & | & 3.32 \0 & 0.068 & 0.909 & | & 0.73\end{bmatrix}\]

Step 3: Create zeros below the second pivot

Continue making zeros below the pivot in the second column (second row, second element). Divide the second row by \(-0.832\) to make the pivot 1, and then create zeros below it by performing: subtract \(0.068/(-0.832)\) times the second row from the third row.This results in:\[\begin{bmatrix}1.1 & 0.7 & -3.1 & | & -1.79 \0 & 1 & -4.234 & | & -3.993 \0 & 0 & 1.196 & | & 0.957\end{bmatrix}\]

Step 4: Back substitution to solve for variables

With the matrix in upper triangular form, perform back substitution.1. Solve for \(z\) in the third row: \ \(1.196z = 0.957 \Rightarrow z = \frac{0.957}{1.196} \approx 0.8\) 2. Substitute \(z\) into the second row to solve for \(y\): \ \(y - 4.234 \cdot 0.8 = -3.993 \Rightarrow y = -3.993 + 4.234 \cdot 0.8 \approx -0.569\)3. Substitute \(y\) and \(z\) into the first row to solve for \(x\): \ \(1.1x + 0.7(-0.569) - 3.1(0.8) = -1.79 \Rightarrow x \approx 1.4\)

Step 5: Final solution

The solutions for the variables based on back substitution are:\[x \approx 1.4, \, y \approx -0.569, \, z \approx 0.8 \]

Key Concepts

Augmented Matrix

To solve a system of equations using Gaussian elimination, we start by converting the system into an augmented matrix. This transformation involves writing the coefficients of each variable and the constants from the equations into a matrix form. It creates a combined structure that makes computations easier for solving the system of equations.
The augmented matrix includes both the variable coefficients and the constants, separated by a division line symbolizing the equals sign in the equations. For example, for the given system, you write it as:

• The first row corresponds to the equation: \(1.1x + 0.7y - 3.1z = -1.79\)
  
• The second row matches: \(2.1x + 0.5y - 1.6z = -0.13\)
  
• The third is: \(0.5x + 0.4y - 0.5z = -0.07\).
  

This matrix-like form simplifies the process of applying elimination techniques, as each manipulation affects an entire equation systematically. Understanding how to interpret and construct this matrix is a foundational skill in linear algebra.

Row Operations

Row operations are key maneuvers in the process of Gaussian elimination. These operations transform the augmented matrix to a simpler form, ideally a triangular one, which makes it easier to solve the system. There are three types of row operations:

• Swapping two rows, if needed, to bring a non-zero number to the top for easier calculations.


• Multiplying a row by a non-zero scalar for finding convenient pivot positions.


• Adding or subtracting rows to eliminate variables systematically below pivot positions.

When performing these operations, it's important to maintain equivalency among the equations. This means the solution set remains unchanged, even though the equations look different now.
For example, in the given exercise, we made all the entries beneath the leading coefficient (pivot) in the first row zero. This was done by strategic subtraction of scalar multiples of the pivot row from subsequent rows. Such operations systematically simplify the matrix towards an upper triangular form.

Back Substitution

After transforming the augmented matrix to an upper triangular matrix through row operations, the next step is back substitution. This technique is used to find the values of the variables from the top down, beginning from the last row of the matrix inward.

Back substitution involves:

• Identifying the last row, which often has an equation in simplified form such as \(1z = c\). Solve for \(z\) directly.


• Substituting the obtained value of \(z\) into the previous row to find \(y\), since it now appears only with known numbers.


• Continuing this pattern upwards, substituting known values to discover \(x\).

This method efficiently pinpoints each variable’s value and is crucial for completing the Gaussian elimination process. The independence of variables after each determination simplifies mathematical work, making it an invaluable part of linear algebra.