Problem 38 - College Algebra

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l}{0.2 x+0.1 y-0.3 z=0.2} \\ {0.8 x+0.4 y-1.2 z=0.1} \\ {1.6 x+0.8 y-2.4 z=0.2}\end{array} $$

Step-by-Step Solution

Answer

The system is inconsistent and has no solutions.

Step 1: Set Up the Augmented Matrix

Write the systems of equations as an augmented matrix. For the given system, the augmented matrix is: \[\begin{bmatrix}0.2 & 0.1 & -0.3 & | & 0.2\0.8 & 0.4 & -1.2 & | & 0.1\1.6 & 0.8 & -2.4 & | & 0.2\end{bmatrix}\]

Step 2: Scale Row 1

To make calculations easier, scale the first row by multiplying it by 5 to avoid decimals:\[\begin{bmatrix}1 & 0.5 & -1.5 & | & 1\0.8 & 0.4 & -1.2 & | & 0.1\1.6 & 0.8 & -2.4 & | & 0.2\end{bmatrix}\]

Step 3: Eliminate x from Row 2

Subtract 0.8 times the new Row 1 from Row 2 to eliminate the x-term:\[Row2 = Row2 - 0.8 \times Row1\]The augmented matrix becomes:\[\begin{bmatrix}1 & 0.5 & -1.5 & | & 1\0 & 0 & 0 & | & -0.7\1.6 & 0.8 & -2.4 & | & 0.2\end{bmatrix}\]

Step 4: Eliminate x from Row 3

Subtract 1.6 times the new Row 1 from Row 3 to eliminate the x-term:\[Row3 = Row3 - 1.6 \times Row1\]The augmented matrix becomes:\[\begin{bmatrix}1 & 0.5 & -1.5 & | & 1\0 & 0 & 0 & | & -0.7\0 & 0 & 0 & | & -1.4\end{bmatrix}\]

Step 5: Analyze the System

Notice that both Row 2 and Row 3 have all-zero coefficients for x, y, and z, but different constants in the augmented part. This indicates a contradiction (inconsistent system), as they imply the equations 0 = -0.7 and 0 = -1.4, which are impossible.

Key Concepts

Understanding the Augmented Matrix

In solving systems of equations like the one given, it's common to use an augmented matrix. This is a way of organizing coefficients and constants into a single matrix for easier manipulation. The augmented matrix combines all the equations' coefficients into rows, and the constants on the right side of the equations form the last column.

Simply put, each row of the augmented matrix represents one equation.

• The coefficients of variables (e.g., numbers before \(x\), \(y\), \(z\)) are arranged in the left columns.
• The constants (the numbers alone on the right of the equal sign) are the final column.

The vertical line separating the last column from the others simply divides the coefficients from the constants. In essence, the augmented matrix is a tool that simplifies row operations and helps us to solve systems effectively.

Identifying an Inconsistent System

An inconsistent system occurs when there is no possible set of solutions that satisfies all the equations simultaneously. In this context, when reducing the augmented matrix using Gaussian elimination, an inconsistent system will show up as a row where the left-hand side equals zero, but the right-hand side does not.

This contradiction means there’s no way the equation can be true. For example:

• If during matrix reduction, you end up with a row such as \(0x + 0y + 0z = -0.7\), the equation says \(0 = -0.7\), which is impossible.
• Similarly, another row such as \(0x + 0y + 0z = -1.4\) implies \(0 = -1.4\), further confirming inconsistency.

Such rows highlight the inconsistency and lead us to conclude that the original set of equations cannot be solved.

How to Eliminate Variables

Eliminating variables is a crucial part of Gaussian elimination. This process simplifies the system of equations by making some variables disappear from certain equations, usually through strategic subtraction or addition.

• The main idea is to transform the system into a triangular form, where the first equation deals with three variables, the second with two, and the last with one.
• In practice, you perform row operations on the matrix, aiming to get zeros in strategic places.
• For instance, to eliminate \(x\) from Row 2, you can subtract a multiple of Row 1 after scaling appropriately to match the coefficient of \(x\) in Row 2.
• This operation involves finding a suitable factor (like 0.8 in the example), multiplying Row 1 by it, and subtracting from Row 2.

By doing so, you systematically remove variables and make the matrix easier to solve, reducing complexity step by step.