Problem 36 - College Algebra with Corequisite Support

Question

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

Step-by-Step Solution

Answer

The system has infinite solutions, with \( x = \frac{24 - 40t - 30s}{60} \), where \( t \) and \( s \) are parameters.

Step 1: Write the system as an augmented matrix

Represent the given system of equations as an augmented matrix. \[\begin{bmatrix}\frac{1}{40} & \frac{1}{60} & \frac{1}{80} & \Big| & \frac{1}{100} \-\frac{1}{2} & -\frac{1}{3} & -\frac{1}{4} & \Big| & -\frac{1}{5} \\frac{3}{8} & \frac{3}{12} & \frac{3}{16} & \Big| & \frac{3}{20}\end{bmatrix}\]

Step 2: Simplify the coefficients

Multiply each equation to eliminate fractions. - First row: Multiply by 2400 (LCM of denominators 40, 60, 80, 100) - Second row: Multiply by 12 (LCM of denominators 2, 3, 4, 5) - Third row: Multiply by 48 (LCM of denominators 8, 12, 16, 20)This gives:\[\begin{bmatrix}60 & 40 & 30 & \Big| & 24 \-6 & -4 & -3 & \Big| & -2.4 \18 & 12 & 9 & \Big| & 7.2\end{bmatrix}\]

Step 3: Perform row operations to obtain upper triangular form

Aim to create zeros below the first pivot (first column).- Multiply the first row by 0.1 and add to the second row.- Multiply the first row by -0.3 and add to the third row.The augmented matrix becomes:\[\begin{bmatrix}60 & 40 & 30 & \Big| & 24 \0 & 0 & 0 & \Big| & 0 \0 & 0 & 0 & \Big| & 0\end{bmatrix}\]

Step 4: Analyze the resulting matrix

Notice that rows 2 and 3 are both zeros, while the first row indicates a valid equation. This implies the system has infinite solutions since there are free variables.

Step 5: Interpret the infinite solutions

Let \( y \) and \( z \) be parameters, say \( y = t \) and \( z = s \). Express \( x \) in terms of \( t \) and \( s \) using the equation obtained from the first row.i.e., \( 60x + 40t + 30s = 24 \)Then, \( x = \frac{24 - 40t - 30s}{60} \).

Key Concepts

Gaussian Elimination

Gaussian elimination is a systematic method used to solve systems of linear equations. This method involves using a series of row operations to simplify a matrix into an upper triangular form, making it easier to deduce the solutions of the system. To begin with Gaussian elimination:

• Rewrite the system of equations in an augmented matrix form.
• Use row operations to zero out entries below the main diagonal.
• Solve the resulting upper triangular matrix through back-substitution.

This technique can help identify not only a unique solution but also situations like infinite solutions or no solutions. In this exercise, Gaussian elimination is employed to discover a condition of infinite solutions due to some rows turning into zero rows during the process.

Augmented Matrix

The augmented matrix is a compact way to represent a system of linear equations. It includes both the coefficient matrix and the constants from the equations' right-hand side, separated by a vertical bar.
For example, if you have a system of three equations, the augmented matrix captures all the essential information needed for solving the system:

• Each row of the matrix corresponds to one equation.
• Except for the constants on the right, each entry before the vertical line represents the coefficients of the variables within that equation.

In the exercise, the given system is first transformed into an augmented matrix form, which simplifies applying row operations efficiently.

Infinite Solutions

A system of linear equations can have infinite solutions if there are at least two variables whose values depend on each other, allowing freedom in their assignment. This scenario typically occurs when the rank of the coefficient matrix is less than the number of variables, leading to one or more free variables.
In this exercise, infinite solutions are indicated by rows in the augmented matrix turning into zero rows after applying Gaussian elimination:

• Since the final simplified matrix has rows of zeros, it implies that those equations are dependent.
• The variables can be expressed in terms of parameters, giving rise to multiple solutions that fulfill the equations.

For our problem, the first equation gives a relation among the variables, meaning any values that satisfy this constraint are solutions.

Matrix Row Operations

Matrix row operations are the steps taken to manipulate rows in a matrix while maintaining the solutions of the system it represents. These operations are crucial to transforming the original augmented matrix into a simpler form:

• Swap two rows to change their positions.
• Multiply a row by a non-zero scalar to scale the equation without changing its solution.
• Add a multiple of one row to another to eliminate variables or create zeros beneath pivots.

By applying these operations smartly, as demonstrated in the solution, we transform the matrix into a form that reveals information about the solution set, such as the presence of free variables and the nature of the solutions.