Problem 40 - College Algebra

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l}{0.5 x-0.5 y+0.5 z=10} \\ {0.2 x-0.2 y+0.2 z=4} \\ {0.1 x-0.1 y+0.1 z=2}\end{array} $$

Step-by-Step Solution

Answer

The system has infinitely many solutions of the form \(x = 20 + y - t\), \(y = y\), \(z = t\).

Step 1: Write the system in augmented matrix form

The given system of equations is: \( 0.5x - 0.5y + 0.5z = 10 \), \( 0.2x - 0.2y + 0.2z = 4 \), \( 0.1x - 0.1y + 0.1z = 2 \). Represent this system in matrix form as an augmented matrix: \[\begin{bmatrix} 0.5 & -0.5 & 0.5 & | & 10 \0.2 & -0.2 & 0.2 & | & 4 \0.1 & -0.1 & 0.1 & | & 2 \end{bmatrix}\]

Step 2: Scale first row to simplify calculations

Divide the entire first row by 0.5 to make calculations simpler and obtain leading coefficient as 1:\[\begin{bmatrix} 1 & -1 & 1 & | & 20 \0.2 & -0.2 & 0.2 & | & 4 \0.1 & -0.1 & 0.1 & | & 2 \end{bmatrix}\]

Step 3: Eliminate first column entries below the pivot

Subtract 0.2 times the first row from the second row and 0.1 times the first row from the third row:\[\begin{bmatrix} 1 & -1 & 1 & | & 20 \0 & 0 & 0 & | & 0 \0 & 0 & 0 & | & 0 \end{bmatrix}\]

Step 4: Draw conclusions from the reduced matrix

Notice the second and third rows are entirely zeros. This indicates that the system has infinitely many solutions, as there are not enough constraints to uniquely determine the values of \(x\), \(y\), and \(z\).

Step 5: Expressing the solutions in parametric form

Since the matrix indicates infinitely many solutions, express dependent variables in terms of a parameter. Let \(z = t\) (where \(t\) is a parameter). From the first equation: \[ x - y + z = 20 \Rightarrow x = 20 + y - z \]Substitute \(z = t\) to get:\[ x = 20 + y - t \]Thus, \(x\), \(y\), and \(z\) can be expressed as: \(x = 20 + y - t\), \(y = y\), \(z = t\).

Key Concepts

System of Equations

A system of equations involves a set of equations with multiple variables that we solve simultaneously. Think of it as finding a common solution for a group of equations. Specifically, we are looking for values of variables that satisfy all the equations at the same time. In the example given, we have three equations:

• 0.5x - 0.5y + 0.5z = 10
• 0.2x - 0.2y + 0.2z = 4
• 0.1x - 0.1y + 0.1z = 2

These equations are linear, meaning they represent straight lines in three-dimensional space. By solving these, we aim to find a point or set of points where these lines intersect, which gives us the solution to the system. The complexity often lies in ensuring all equations are satisfied simultaneously, which requires a systematic approach like Gaussian elimination.

Augmented Matrix

An augmented matrix is a convenient way to handle a system of linear equations. It compiles the coefficients of the variables and the constants from each equation into a single matrix. For the given problem, the system is first written in matrix form:\[\begin{bmatrix}0.5 & -0.5 & 0.5 & | & 10 \0.2 & -0.2 & 0.2 & | & 4 \0.1 & -0.1 & 0.1 & | & 2 \\end{bmatrix}\]Here, each row represents one equation, and the columns on the left of the vertical line represent the coefficients of the variables \(x\), \(y\), and \(z\). The column on the right of the vertical line includes the constants from the equations. The vertical line itself symbolizes the equals sign in the original equations. This format makes it easier to apply mathematical operations to simplify and solve the system, allowing us to focus on the manipulation of numbers rather than the equations themselves.

Parametric Form

When a system of equations has infinitely many solutions, we use parametric form to express these solutions. Parametric form expresses the solution set in terms of one or more free variables, also known as parameters. For the example provided, after performing Gaussian elimination, the matrix is reduced to:\[\begin{bmatrix}1 & -1 & 1 & | & 20 \0 & 0 & 0 & | & 0 \0 & 0 & 0 & | & 0 \\end{bmatrix}\]The zero rows imply there are free variables, meaning there aren't enough unique equations to determine all variable values independently. To accommodate this, a parameter \(t\) is introduced, often related to one variable, typically \(z\) in this case. We express other variables in terms of \(t\):

• Let \(z = t\)
• From first equation: \(x - y + z = 20\) implies \(x = 20 + y - t\)
• Thus, \(y\) remains a free variable

Hence, the solutions are parameterized as: \((x, y, z) = (20 + y - t, y, t)\). This expression allows us to understand the set of all possible solutions for our system.