Problem 28 - A Graphical Approach to Precalculus with Limits

Question

Each augmented matrix is in row echelon form and represents a linear system. Use back-substitution to solve the system if possible. $$\left[\begin{array}{rrr|r} 1 & -1 & 2 & 8 \\ 0 & 1 & -4 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution set is \((x_1, x_2, x_3) = (6t + 10, 4t + 2, t)\) for any real \(t\).

Step 1: Analyze the Augmented Matrix

The given augmented matrix is \( \begin{bmatrix} 1 & -1 & 2 & 8 \ 0 & 1 & -4 & 2 \ 0 & 0 & 0 & 0 \end{bmatrix} \). Notice that the third row is composed entirely of zeros, which indicates a dependent system with fewer equations than variables.

Step 2: Set Up the Corresponding System of Equations

Translate the non-zero rows of the matrix into equations. The first row, \([1, -1, 2 | 8]\), translates to the equation \(x_1 - x_2 + 2x_3 = 8\). The second row, \([0, 1, -4 | 2]\), translates to \(x_2 - 4x_3 = 2\).

Step 3: Express the Leading Variables

Solve for the leading variables. From the second equation, express \(x_2\) in terms of \(x_3\): \(x_2 = 4x_3 + 2\). Substitute \(x_2\) into the first equation to express \(x_1\): \(x_1 = x_2 - 2x_3 + 8\). Simplifying, \(x_1 = 6x_3 + 10\).

Step 4: Write the General Solution

Since the third row was all zeros, \(x_3\) is a free variable. Express \(x_1\) and \(x_2\) in terms of \(x_3\), the parameter. The solution is: \(x_1 = 6x_3 + 10\), \(x_2 = 4x_3 + 2\), and \(x_3 = x_3\). Thus, the general solution in parametric form is \( (x_1, x_2, x_3) = (6t + 10, 4t + 2, t) \) where \(t\) is any real number.

Key Concepts

Row Echelon Form

In linear algebra, one of the key methods used to express linear systems is the Row Echelon Form (REF). When a matrix is in this form, it simplifies solving the corresponding system of equations. A matrix is considered to be in row echelon form when these rules are met:

• All non-zero rows are above any rows of all zeroes.
• The leading entry (also called a pivot) of each non-zero row is to the right of the leading entry of the row above it.
• The leading entry in any row is 1, if possible.

For example, the given matrix, \[\begin{bmatrix} 1 & -1 & 2 & 8 \ 0 & 1 & -4 & 2 \ 0 & 0 & 0 & 0 \end{bmatrix}\]fits these criteria, making it a row echelon form. The matrix showcases a dependent system, as the third row is all zeros, suggesting fewer equations than variables. This sets the stage for solving the system using techniques like back-substitution.

Back-Substitution

Back-substitution is a method used to solve systems of equations that have been converted to row echelon form. It involves starting from the last non-zero row and working upwards to find the variable values. In our case, once the matrix is already in REF, we translate it to the system of equations:

• \(x_1 - x_2 + 2x_3 = 8\)
• \(x_2 - 4x_3 = 2\)

Since the third row is zero, it tells us there is potential for a free variable. From the second equation, we solve for \(x_2\) in terms of \(x_3\): \[x_2 = 4x_3 + 2\] We then substitute \(x_2\) in the first equation to express \(x_1\): \[x_1 = 6x_3 + 10\] This provides the solution in terms of a single parameter \(t\), allowing us to write the answer in parametric form.

Parametric Equations

Parametric equations offer a convenient way to express solutions of systems with free variables. In this context, the solution of a linear system may not be unique. Instead, a whole set of solutions expressed using a parameter could be found. In our exercise, \(x_3\) is taken as the parameter because the corresponding row in the matrix contained all zeros.By solving:

• \(x_1 = 6x_3 + 10\)
• \(x_2 = 4x_3 + 2\)
• \(x_3 = x_3\)

We incorporate \(x_3\) as the parameter \(t\), leading to the parametric form:\[(x_1, x_2, x_3) = (6t + 10, 4t + 2, t)\]This representation communicates that each variable depends on \(t\), a real number, thus describing an infinite set of solutions. Parametric solutions are particularly useful for systems with infinitely many possible solutions, providing a way to easily write all of them.