Problem 48 - College Algebra and Calculus: An Applied Approach

Question

Write the system of equations represented by the augmented matrix. Use backsubstitution to find the solution. (Use \(x, y, z\), and \(w .)\) $$ \left[\begin{array}{rrrrrr} 1 & 2 & -2 & 0 & \vdots & -1 \\ 0 & 1 & 1 & 2 & \vdots & 9 \\ 0 & 0 & 1 & 0 & \vdots & 2 \\ 0 & 0 & 0 & 1 & \vdots & -3 \end{array}\right] $$

Step-by-Step Solution

Answer

The solution of the system represented by the given augmented matrix is \((-23, 13, 2, -3)\).

Step 1: Write system of equations from the augmented matrix

The given augmented matrix represents a system of equations. Each row in the matrix corresponds to an equation. The coefficients in the first four columns become the coefficients of the variables \(x, y, z, w\), and the last column represents the constants on the right side of the equal sign. Therefore, the system of equations represented by the augmented matrix is: \[ \begin{cases} x + 2y - 2z = -1 \ y + z + 2w = 9 \ z = 2 \ w = -3 \end{cases} \]

Step 2: Use back-substitution to solve for \(x, y, z, w\)

We start the back-substitution with the last two equations as they have single variables. We get \(z=2\) and \(w=-3\) from the third and fourth equation. Now substituting the values of \(z\) and \(w\) in the second equation: \[ y + z + 2w = y + 2 - 6 = y - 4 = 9 \]So, \(y = 13\). Now, substituting \(y=13\) and \(z=2\) in the first equation to find \(x\): \[ x = -1 - 2y + 2z = -1 - 2*13 + 2*2 = -1 - 26 + 4 = -23 \] So, \(x = -23\)

Step 3: Write down the solution set

We now have the values of all our variables. Thus, the solution to the system is: \[ (x, y, z, w) = (-23, 13, 2, -3) \]

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of unknowns. In this exercise, we are looking at a system involving four variables: \(x\), \(y\), \(z\), and \(w\). Each equation corresponds to one row of the given augmented matrix, making it easy to visualize the problem. This matrix format is often used in linear algebra to efficiently solve systems of equations.

Let's break it down to understand it more clearly:

• The coefficients of the variables \(x, y, z,\) and \(w\) are derived from the entries in the first four columns of the matrix.
• The last column of the matrix consists of constants that appear on the right side of each equation's equal sign.

This structured approach allows us to systematically solve for multiple variables step by step. Here, each equation is interlinked, guiding us to the solution when the equations are solved together.

Back-Substitution

Back-substitution is a technique used to find solutions in systems of equations, especially when the system is represented in an upper triangular form (like our matrix). Simply put, it involves solving the equations from the bottom up, starting with the last equation.

So how does it work?

• Begin with the last equation: because it often involves only one variable, it's easy to solve directly.
• Substitute the known values back into the previous equations, simplifying them step by step. Each time you insert a known variable, it reduces the number of unknowns in the next equation.
• Continue the process until all variables are solved for.

In our example, we tackled the system using back-substitution, working from the given \(z = 2\) and \(w = -3\), which helped us find \(y = 13\) and finally \(x = -23\). By systematically plugging the known values back into other equations, we narrowed down on the final solution set.

Linear Algebra

Linear algebra is a significant branch of mathematics that deals with vectors, matrices, and linear transformations. The solution of a system of equations using augmented matrices is a fundamental concept in linear algebra.

Here's why linear algebra is important:

• It provides a systematic approach for solving systems with multiple linear equations.
• Matrices, like the one given in the exercise, help simplify complex sets of linear equations into a manageable format. This helps streamline calculations and solutions.
• Concepts in linear algebra are crucial in fields such as engineering, computer science, physics, and more due to their ability to model complex systems.

By mastering the basics of linear algebra, such as understanding matrices and systems of equations, you can handle larger, more complex operations with confidence. This exercise serves as a building block to tackle larger mathematical challenges.