Problem 37 - Contemporary Precalculus

Question

Exercises \(37-40,\) solve the system. [Note: The REF and RREF keys on some calculators produce an error message when there are more rows than columns in a matrix, in which case you will have to solve the system by some other means.] $$\begin{aligned} &2 x-y=1\\\ &3 x+y=2\\\ &4 x-2 y=2\\\ &5 x+5 y=4 \end{aligned}$$

Step-by-Step Solution

Answer

Answer: The values of x and y in the given system of linear equations are x = 2/5 and y = 1/5.

Step 1: Write the augmented matrix of the system

First, we need to write the augmented matrix corresponding to the given system, which will have the coefficients of the variables as the elements of the matrix. The augmented matrix will be represented as follows: $$ \left[ \begin{array}{ccc|c} 2 & -1 & 1 \\ 3 & 1 & 2 \\ 4 & -2 & 2 \\ 5 & 5 & 4 \end{array} \right] $$

Step 2: Perform Row Operations

Now, we will perform row operations to transform the matrix into an upper-triangular form. Start by eliminating the variables in the second, third, and fourth rows. We can do this by adding and subtracting appropriate multiples of the first row from the other rows. Row operation: R₂ - 1.5R₁ -> R₂ $$ \left[ \begin{array}{ccc|c} 2 & -1 & 1 \\ 0 & 2.5 & 0.5 \\ 4 & -2 & 2 \\ 5 & 5 & 4 \end{array} \right] $$ Row operation: R₃ - 2R₁ -> R₃ $$ \left[ \begin{array}{ccc|c} 2 & -1 & 1 \\ 0 & 2.5 & 0.5 \\ 0 & 0 & 0 \\ 5 & 5 & 4 \end{array} \right] $$ Row operation: R₄ - 2.5R₁ -> R₄ $$ \left[ \begin{array}{ccc|c} 2 & -1 & 1 \\ 0 & 2.5 & 0.5 \\ 0 & 0 & 0 \\ 0 & 7.5 & 1.5 \end{array} \right] $$ Next, eliminate the variable in the third row. Row operation: R₄ - 3R₂ -> R₄ $$ \left[ \begin{array}{ccc|c} 2 & -1 & 1 \\ 0 & 2.5 & 0.5 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$

Step 3: Solve for the variables

Now, our matrix is in upper-triangular form. We can notice that the third and fourth equations have become trivial, implying that there is a dependent system, meaning the equations aren't independent and one of them is redundant. Based on the non-trivial equations (1 and 2), we can obtain the solutions. First, solve for y from the second equation: $$ y = \frac{0.5}{2.5} = \frac{1}{5} $$ Now, plug the value of y into the first equation and solve for x: $$ 2x - \left(-\frac{1}{5}\right) = 1 $$ $$ 2x + \frac{1}{5} = 1 $$ $$ 2x = \frac{4}{5} $$ $$ x = \frac{2}{5} $$ So, the solution to the system of equations is: $$ x = \frac{2}{5}, \qquad y = \frac{1}{5} $$

Key Concepts

Augmented Matrix

An augmented matrix is a compact way of combining the coefficients and constant terms of a linear system of equations. It's like a shorthand for your system that helps you work more comfortably, especially when performing operations to solve the system.

In the context of the problem, the augmented matrix is:

• The left side, consisting of coefficients of variables (\[\left[\begin{array}{cc}2 & -1 \3 & 1 \4 & -2 \5 & 5\end{array}\right]\]) represents the variables in the equations.
• The rightmost column (\[\left[\begin{array}{c}1 \2 \2 \4\end{array}\right]\]) is the result from each equation.

Writing it this way prepares us to apply row operations, streamlining the process of solving the system.

Row Operations

Row operations are the steps we take to modify a matrix, making it easier to read and solve the system it represents. These operations help convert a matrix into an upper-triangular or row-reduced form, which simplifies finding solutions.

• Adding or subtracting complete rows: This helps eliminate coefficients beneath the pivot positions, turning them into zeros.
• Multiplying a row by a non-zero scalar: Necessary for clarity and can simplify equations.
• Swapping rows: Sometimes used to position pivot elements strategically.

For example, in the original exercise, row operations are used to methodically eliminate x and y terms from all but the first equation, culminating in a simpler equation set. This is crucial for identifying when systems are dependent or independent.

Dependent System

A dependent system arises when one equation in the system can be expressed as a combination of the others. This means not all of the equations provide new information, leading to multiple solutions or no unique solutions.

In our exercise, once row operations are completed, we find that the last two rows of the matrix are entirely zeros:

• This indicates that these equations do not add new information, often meaning they are combinations of the others.
• Such systems generally imply reliance among the equations rather than independence.

Understanding dependent systems is integral to solving them efficiently. When identified, focus shifts to the significant equations within the system, seeking solutions only from them. It saves computational time and fosters deeper comprehension of system dynamics.