Problem 40 - College Algebra with Corequisite Support

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

Infinitely many solutions; the general solution can be parameterized.

Step 1: Write the augmented matrix

First, we need to represent the given system of equations as an augmented matrix. The system is: \( \begin{align*} 0.5x - 0.5y + 0.5z &= 10 \ 0.2x - 0.2y + 0.2z &= 4 \ 0.1x - 0.1y + 0.1z &= 2 \end{align*} \). The augmented matrix is: \[ \begin{bmatrix} 0.5 & -0.5 & 0.5 & | & 10 \ 0.2 & -0.2 & 0.2 & | & 4 \ 0.1 & -0.1 & 0.1 & | & 2 \end{bmatrix} \].

Step 2: Scale the first row

To make calculations easier, we scale the first row by multiplying it by 2. The first row becomes \[ \begin{bmatrix} 1 & -1 & 1 & | & 20 \end{bmatrix} \]. Now the matrix is: \[ \begin{bmatrix} 1 & -1 & 1 & | & 20 \ 0.2 & -0.2 & 0.2 & | & 4 \ 0.1 & -0.1 & 0.1 & | & 2 \end{bmatrix} \].

Step 3: Eliminate x from the second and third rows

Use the first row to eliminate \(x\) from the second and third rows by substituting: - For the second row: Multiply row 1 by 0.2 and subtract from row 2.- For the third row: Multiply row 1 by 0.1 and subtract from row 3.The operations yield: \[ \begin{bmatrix} 1 & -1 & 1 & | & 20 \ 0 & 0 & 0 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \].

Step 4: Analyze the reduced matrix

The reduced matrix \( \begin{bmatrix} 1 & -1 & 1 & | & 20 \ 0 & 0 & 0 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \) suggests that there are infinitely many solutions since there are not enough equations to uniquely determine the variables. Solving \(x - y + z = 20\) allows us to express one variable in terms of others and parameterize the solution.

Step 5: Express the solution in parametric form

Choose a parameter, such as \(z = t\). Substituting back into our only equation, we have: \[ x - y + t = 20 \] Choose another parameter for \(y\), say \(y = s\). Then \(x = s - t + 20\). The parametric solutions are: \( x = t + 20 \), \( y = t \), \( z = t \).

Key Concepts

Understanding Augmented Matrices

In the realm of linear algebra, an augmented matrix is a powerful tool that simplifies solving systems of linear equations. It is a combination of the coefficient matrix with the constant terms from each equation added as a final column. This transformation allows us to use matrix operations to find solutions systematically.

Consider the system of equations from the exercise:

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

To write this as an augmented matrix, we align all coefficients and constant terms as follows:\[\begin{bmatrix}0.5 & -0.5 & 0.5 & | & 10 \0.2 & -0.2 & 0.2 & | & 4 \0.1 & -0.1 & 0.1 & | & 2\end{bmatrix}\]This matrix form helps in performing row operations to simplify and ultimately solve the equations. By consecutively eliminating variables, often using the Gaussian elimination method, you can arrive at a simpler form from which solutions can be easily extracted.

Nature of Infinitely Many Solutions

Infinitely many solutions arise in systems of equations when there are fewer unique equations than variables, leading to dependent equations. In this exercise, by reducing the matrix you end with rows full of zeros:\[\begin{bmatrix}1 & -1 & 1 & | & 20 \0 & 0 & 0 & | & 0 \0 & 0 & 0 & | & 0\end{bmatrix}\]This indicates a rank deficiency, meaning the system doesn't have a unique solution. Instead, it lays on a line or plane of solutions that extend infinitely. The leading row (non-zero row) equation is:\[x - y + z = 20\]This does not uniquely determine the values of \(x\), \(y\), and \(z\), allowing for at least one variable to assume infinitely many values, defined as other variables take on any number of values as well.

Explaining Parametric Form

To express solutions of a system with infinitely many solutions, we use parametric form. This involves expressing each variable in terms of one or more parameters. Let’s continue with our simplified equation from before:\[x - y + z = 20\]A parameter, represented by \(t\), is chosen for one variable, for instance, \(z = t\). Substituting back, we find relationships:

• By setting \(y = s\), we can express: \(x = 20 + t + s\)
• Choose another variable, \(z\) as a parameter: \(z = t\)
• And \(y = s\)

Thus, the broader solution in parametric form becomes:

• \(x = s - t + 20\)
• \(y = s\)
• \(z = t\)

The parameters \(s\) and \(t\) can be any real numbers, demonstrating an infinite number of solutions adhering to the conditions of the original equation. This representation captures the essence of infinite solutions an equation can have.