Problem 18 - College Algebra

Question

In Exercises \(13-18\), write the system of linear equations represented by the augmented matrix. Use \(x, y, z\) and, if necessary, \(w, x, y,\) and \(z,\) for the variables. Once the system is written, use back substitution to find its solution. \(\left[\begin{array}{rrrr|r}1 & 2 & -1 & 0 & 2 \\ 0 & 1 & 1 & -2 & -3 \\ 0 & 0 & 1 & -1 & -2 \\ 0 & 0 & 0 & 1 & 3\end{array}\right]\)

Step-by-Step Solution

Answer

The solution of the system of equations given by the augmented matrix is \((x, y, z, w) = (-1, 2, 1, 3)\).

Step 1: Writing the System from the Augmented Matrix

First, write the system of linear equations that the given augmented matrix represents. The augmented matrix given is: \[\left[\begin{array}{rrrr|r}1 & 2 & -1 & 0 & 2 \\ 0 & 1 & 1 & -2 & -3 \\ 0 & 0 & 1 & -1 & -2 \\ 0 & 0 & 0 & 1 & 3\end{array}\right]\]. This matrix can be written as the following system of linear equations:\[\begin{aligned} x + 2y - z &= 2, \\ y+ z - 2w &= -3, \\ z - w &= -2, \\ w &= 3. \end{aligned}\]

Step 2: Back Substitution

In the next step, to find the solution to this system of linear equations, back substitution is done. Start with the bottom equation, the solution for \(w\) is directly given as 3. The third equation gives \(z = w - 2 = 3 - 2 = 1\). The second equation gives \(y = -3 - z + 2w = -3 - 1 + 6 = 2\). Finally, the first equation gives \(x = 2 - 2y + z = 2 - 4 + 1 = -1\).

Step 3: Solution

Therefore, the solution of the system of equations given by the augmented matrix is \((x, y, z, w) = (-1, 2, 1, 3)\).

Key Concepts

Augmented Matrix

An augmented matrix is a highly efficient way to represent a system of linear equations for solving. It includes all the coefficients of the variables and the constants from the equations lined up in a rectangular array. An augmented matrix is split into two parts by a vertical bar: the left side shows the coefficients of the variables, while the right side shows the constants.

For example, consider the matrix from the exercise:
\[\[\begin{align*}\left[\begin{array}{cccc|c}1 & 2 & -1 & 0 & 2 \0 & 1 & 1 & -2 & -3 \0 & 0 & 1 & -1 & -2 \0 & 0 & 0 & 1 & 3\end{array}\right]\end{align*}\]\]
This represents a system where each row corresponds to an equation and each column (before the bar) represents a variable (in the order of x, y, z, w). The numbers to the right of the bar are the constants from each equation. This compact presentation is not only neat but paves the way for methods like Gaussian elimination and back substitution to solve the equations efficiently.

Back Substitution

The method of back substitution is used after transforming the system of equations into an upper triangular form, where all entries below the main diagonal are zero. Starting from the last equation, you find the value of the last variable, and then you substitute this back into the preceding equation to find the second last variable, and so on.

In our exercise, the augmentation has already been handled to form an upper triangular matrix. Starting from the last row, we-determine the value for w, and then use that value in the third row to find z, followed by the second and first rows to find y and x, respectively:
- w is determined to be 3 from the last row.
- Substituting w into the third row, we find that z is 1.
- Continuing in this fashion, y is found to be 2, and finally, x is -1.

This methodical approach ensures that each step is simple arithmetic involving only one unknown at a time, making calculations straightforward and reducing the chances of making errors.

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. Solving a system of linear equations means finding the values of the variables that satisfy all equations simultaneously.

The system derived from our original augmented matrix is expressed as:
\[\[\begin{align*}x + 2y - z &= 2, \ y + z - 2w &= -3, \ z - w &= -2, \ w &= 3.\end{align*}\]\]

System solutions can be categorized as:

• Unique solution: One set of values for the variables that solves all equations.
• No solution: No set of values exists that can satisfy all equations simultaneously (the equations are inconsistent).
• Infinitely many solutions: There is more than one set of values that satisfies all equations, typically when the system has fewer independent equations than variables.

In this case, we are given a unique solution where each equation adds a piece of information to restrict the values of variables, leading to one precise solution.

Algebraic Solution

The algebraic solution to a system of linear equations involves manipulating the equations algebraically to find the values of the variables. Back substitution, which was executed in this exercise, is one such algebraic method. Other methods include substitution, elimination, and graphical methods.

The algebraic solution found for the system from the augmented matrix is
(\(-1\text{, }2\text{, }1\text{, }3\)). This means that when x is -1, y is 2, z is 1, and w is 3, all four equations are satisfied simultaneously. Algebraic solutions require careful step-by-step manipulations. Each operation performed should be valid algebraically to ensure the integrity of the solution is maintained throughout the process.

Understanding different methods for finding an algebraic solution is valuable, as some can be more efficient or easier to apply depending on the particular system of equations at hand.