Problem 43 - College Algebra

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l}{0.5 x+0.2 y-0.3 z=1} \\ {0.4 x-0.6 y+0.7 z=0.8} \\ {0.3 x-0.1 y-0.9 z=0.6}\end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{729}{697}, y = \frac{13}{38}, z = -\frac{1}{697}\).

Step 1: Write the Augmented Matrix

First, convert the given system of equations into an augmented matrix. The system is:\[\begin{array}{l} {0.5 x+0.2 y-0.3 z=1} \ {0.4 x-0.6 y+0.7 z=0.8} \ {0.3 x-0.1 y-0.9 z=0.6} \end{array}\]The augmented matrix is:\[\begin{bmatrix}0.5 & 0.2 & -0.3 & | & 1 \0.4 & -0.6 & 0.7 & | & 0.8 \0.3 & -0.1 & -0.9 & | & 0.6\end{bmatrix}\]

Step 2: Make the First Column a Leading 1

Divide the first row by 0.5 to make the first element of the first row (the pivot) equal to 1. The first row becomes:\[1 & 0.4 & -0.6 & | & 2\]The matrix now looks like:\[\begin{bmatrix}1 & 0.4 & -0.6 & | & 2 \0.4 & -0.6 & 0.7 & | & 0.8 \0.3 & -0.1 & -0.9 & | & 0.6\end{bmatrix}\]

Step 3: Use Row Operations to Create Zeros Below the Pivot

Subtract 0.4 times the first row from the second row to make the first element of the second row zero:\[R2 = R2 - 0.4R1 \]This results in:\[0 & -0.76 & 0.94 & | & -0.02\]Subtract 0.3 times the first row from the third row to eliminate the first element of the third row:\[R3 = R3 - 0.3R1\]This results in:\[0 & -0.22 & -0.72 & | & 0\]The new matrix is:\[\begin{bmatrix}1 & 0.4 & -0.6 & | & 2 \0 & -0.76 & 0.94 & | & -0.02 \0 & -0.22 & -0.72 & | & 0\end{bmatrix}\]

Step 4: Make the Second Column a Leading 1

Divide the second row by -0.76 to make the second pivot 1:\[R2 = \frac{1}{-0.76} R2\]This results in:\[0 & 1 & -\frac{47}{38} & | & \frac{1}{38}\]The matrix is:\[\begin{bmatrix}1 & 0.4 & -0.6 & | & 2 \0 & 1 & -\frac{47}{38} & | & \frac{1}{38} \0 & -0.22 & -0.72 & | & 0\end{bmatrix}\]

Step 5: Use Row Operations to Create Zeros in the Second Column Below the Pivot

Add 0.22 times the second row to the third row to eliminate the second element of the third row:\[R3 = R3 + 0.22R2\]This results in:\[0 & 0 & -\frac{697}{475} & | & \frac{1}{475}\]The new matrix is:\[\begin{bmatrix}1 & 0.4 & -0.6 & | & 2 \0 & 1 & -\frac{47}{38} & | & \frac{1}{38} \0 & 0 & -\frac{697}{475} & | & \frac{1}{475}\end{bmatrix}\]

Step 6: Solve for the Variables

Back substitution begins from the last row. Solve for \( z \) using the third row:\[-\frac{697}{475}z = \frac{1}{475}\]Thus:\[ z = -\frac{1}{697} \]Substitute \( z \) into the second row to solve for \( y \):\[1y - \frac{47}{38}(-\frac{1}{697}) = \frac{1}{38}\]yields:\[ y = \frac{13}{38}\]Substitute both \( y \) and \( z \) into the first row to solve for \( x \):\[1x + 0.4(\frac{13}{38}) - 0.6(-\frac{1}{697}) = 2\]yields:\[ x = \frac{729}{697}\]

Step 7: Present the Solution

The solution to the system of equations is:\[x = \frac{729}{697}, \quad y = \frac{13}{38}, \quad z = -\frac{1}{697}\]

Key Concepts

Augmented Matrix

In order to solve a system of linear equations using Gaussian elimination, we first need to represent the system in an organized form called an augmented matrix. An augmented matrix combines the coefficients of the variables from the linear equations and their corresponding constant terms in one rectangular array.

For instance, given the system:
0.5x + 0.2y - 0.3z = 1,
0.4x - 0.6y + 0.7z = 0.8,
0.3x - 0.1y - 0.9z = 0.6,

we put them into a matrix that looks like this: \[\begin{bmatrix}0.5 & 0.2 & -0.3 & | & 1 \0.4 & -0.6 & 0.7 & | & 0.8 \0.3 & -0.1 & -0.9 & | & 0.6\end{bmatrix} \]This matrix combines all the information from the equations in a neat format. The vertical bar separates the coefficients of the variables from the constants, clearly delineating the two parts of the equation.

This setup is the starting point of Gaussian elimination, simplifying the process of using row operations to manipulate the system into a form that is easy to solve.

Row Operations

Row operations are crucial techniques used in Gaussian elimination to manipulate the augmented matrix. These operations aim to simplify the matrix, making it easier to find solutions to the system of linear equations. There are three types of row operations we can perform:

• Row swapping: We can interchange two rows at any time. This may help move a row with a better pivot or leading coefficient into a particular position.
• Scalar multiplication: Multiply all entries in a row by a non-zero scalar to create leading coefficients of 1 (known as pivots).
• Row addition/subtraction: Add or subtract a multiple of one row to another to create zeros below or above pivots.

Using these operations, the goal is to transform the matrix into a reduced row-echelon form. This is akin to stair-stepping the matrix, where leading variables become apparent, enabling easy back substitution. For example, in our problem, creating zeros below the first pivot simplifies the equations to focus on the interactions between the remaining variables.

Back Substitution

After transforming the matrix into an upper triangular form using row operations, we use back substitution to find the solutions to the system. Back substitution involves solving for the variables starting from the bottom row of the matrix and working upwards.

Here's how it works: Once we reach the last row, it represents an equation with a single variable. Solving for this variable first, we then substitute it back into the previous rows. This process is repeated until all variables are solved, working our way up.

Take the third row in the reduced matrix, which simplifies to:
\[-\frac{697}{475}z = \frac{1}{475}\] Solving for \(z\), we then substitute \(z\) back into the second equation, followed by the first, to determine \(y\) and then \(x\). This systematic unraveling ensures each variable is accurately calculated in terms of those already found, thereby avoiding errors and verifying the steps taken.

System of Linear Equations

A system of linear equations consists of two or more linear equations with shared variables. These are called simultaneous equations because we want to find values for the variables that satisfy all equations at the same time.

Such systems can model many real-world phenomena, like finding equilibrium points, optimizing resources, or analyzing networks. For example, equations in the form of:
0.5x + 0.2y - 0.3z = 1,
0.4x - 0.6y + 0.7z = 0.8,
0.3x - 0.1y - 0.9z = 0.6

represent a typical situation where multiple conditions or constraints intersect. The main goal is to determine the exact values of \(x, y, \) and \(z\) that satisfy each equation simultaneously.

Solving these systems involves methods like substitution, elimination, and more systematically, matrix operations such as Gaussian elimination. This method is particularly powerful as it can handle systems with many equations more efficiently than solving them one by one.