Problem 36 - College Algebra

Question

For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{l}{\frac{1}{40} x+\frac{1}{60} y+\frac{1}{80} z=\frac{1}{100}} \\\ {-\frac{1}{2} x-\frac{1}{3} y-\frac{1}{4} z=-\frac{1}{5}} \\ {\frac{3}{8} x+\frac{3}{12} y+\frac{3}{16} z=\frac{3}{20}}\end{array} $$

Step-by-Step Solution

Answer

The system has infinite solutions: \[ x = \frac{4}{5}z - \frac{4}{5}, \; y = \frac{3}{10}z - \frac{3}{10}, \; z = z \].

Step 1: Write the Augmented Matrix

First, let's write the system of equations in an augmented matrix form: \[\begin{bmatrix} \frac{1}{40} & \frac{1}{60} & \frac{1}{80} & | & \frac{1}{100} \-\frac{1}{2} & -\frac{1}{3} & -\frac{1}{4} & | & -\frac{1}{5} \\frac{3}{8} & \frac{3}{12} & \frac{3}{16} & | & \frac{3}{20} \end{bmatrix} \]

Step 2: Clear Fractions

To simplify the calculations, multiply each row by the least common multiple of the denominators to clear fractions:1. Multiply the first row by 2400.2. Multiply the second row by 12.3. Multiply the third row by 48.The augmented matrix becomes:\[\begin{bmatrix} 60 & 40 & 30 & | & 24 \-6 & -4 & -3 & | & -2 \18 & 12 & 9 & | & 7.2 \end{bmatrix} \]

Step 3: Perform Row Operations (Row 1)

Begin by making the leading coefficient of Row 1, 1 if it was not already, and clear the first column of other rows using row operations. Here Row 1 can be used to eliminate the x-terms in Rows 2 and 3.1. Replace Row 2 with Row 2 + (1/10) * Row 1.2. Replace Row 3 with Row 3 - (3/10) * Row 1.The augmented matrix becomes:\[\begin{bmatrix} 60 & 40 & 30 & | & 24 \0 & 0 & 0 & | & 0 \0 & 0 & 0 & | & 0 \end{bmatrix} \]

Step 4: Analyze the System

The matrix achieved in the previous step shows that Rows 2 and 3 have all their coefficients as zero, implying that they do not contribute any new information to the system. Thus, the system has possibly infinite solutions. To further confirm, you can solve the remaining equation: \[60x + 40y + 30z = 24.\]

Step 5: Express Variables in Terms of One Variable

Let's decide to express all variables in terms of one variable, say \( z \).Re-arrange the equation from Row 1:\[60x + 40y + 30z = 24 \]Divide by 30:\[2x + \frac{4}{3}y = \frac{4}{5} - z \]So, express \(x\) and \(y\) in terms of \(z\).\[x = \frac{4}{5}z - \frac{4}{5} \]\[y = \frac{3}{10}z - \frac{3}{10} \]

Step 6: General Solution

Conclude with a general solution:\[x = \frac{4}{5}z - \frac{4}{5}, \quad y = \frac{3}{10}z - \frac{3}{10}, \quad z = z.\]This is the parametric form of the solution space where \( z \) can be any real number.

Key Concepts

Linear Systems

Linear systems consist of multiple linear equations with variables that we need to solve simultaneously. In general, a linear equation is expressed in the form of ax + by + cz = d, where a, b, and c are coefficients and x, y, z are the unknowns.
When dealing with a system of equations, we have multiple such lines that we want to solve for identical x, y, z.
This is common in real-world problems where multiple constraints need to be satisfied.
A solution to a linear system can be:

• **Unique**: Exactly one set of values for the unknowns satisfies all equations simultaneously.
• **Infinite**: There are an infinite number of sets of values that satisfy the equations, often represented parametrically.
• **No solution**: No values exist that satisfy all equations, often when the lines are parallel and do not intersect.

Solving linear systems efficiently is crucial in many areas like engineering, physics, and economics.

Augmented Matrix

An augmented matrix is a compact way of representing a system of linear equations. It brings the equations together into a single matrix form, making it easier to apply operations systematically.
In an augmented matrix, the coefficients of the variables form the matrix on the left side, while the constants are added as a column vector on the right side, separated by a vertical line.
For example, for linear equations:

• \( ax + by + cz = d \)
• \( ex + fy + gz = h \)

The augmented matrix representation would be:",