Problem 94 - Precalculus with Limits

Question

In Exercises 91-94, determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices. (a) \( \left\\{ \begin{array}{l} x + 3y - z = 19 \\ y + 6z = -18 \\ z = -4 \\ \end{array} \right. \) (b) \( \left\\{ \begin{array}{l} x - y + 3z = -15 \\ y - 2z = 14 \\ z = -4 \\ \end{array} \right. \)

Step-by-Step Solution

Answer

Yes, the two systems yield the same solution, which is \(x=1\), \(y=6\), and \(z=-4\).

Step 1: Find the solution for the first system

Starting with the third equation, we have \( z = -4 \). Now, substitute this into the second equation to find \(y\), which gives \(y = -18 - 6*(-4) = 6\). Finally, substitute these into the first equation to find \(x\), which gives \(x = 19 - 3*6 + 4 = 1\). So, the solution to the system is \(x=1\), \(y=6\), and \(z=-4\).

Step 2: Find the solution for the second system

Similarly, starting with the third equation, we still have \( z = -4 \). Now, substitute this into the second equation to find \(y\), which gives \(y = 14 + 2*(-4) = 6\). Finally, substitute these into the first equation to find \(x\), which gives \(x = -15 + 6 - 3*(-4) = 1\). Thus, the solution to this system is also \(x=1\), \(y=6\), and \(z=-4\).

Step 3: Compare the solutions

Since both systems have the same solution, \(x=1\), \(y=6\), and \(z=-4\), we can safely conclude that the two systems of linear equations yield the same solution.

Key Concepts

Solving Systems Using Matrices

When faced with systems of linear equations, matrices can be a powerful tool for finding solutions efficiently. A matrix is essentially a rectangular array of numbers arranged in rows and columns that can represent a system's coefficients. By transforming a system into matrix form, we can apply various techniques, such as Gaussian elimination or using the inverse of a matrix, to solve for the variables.

Consider our exercise, where solving each system can be streamlined by creating a matrix A with the coefficients of the variables and a corresponding matrix B with the constants from the right side of the equations. We then seek a solution in the form of matrix X, which represents our variables such that AX = B. For a 3x3 system, we might see this method take the form of reducing the augmented matrix [A|B] to row-echelon form, making it easier to find the solution for each variable sequentially.

This process is valuable as it can be used for systems of any size, making it an ideal method for complex or larger systems. Understanding how to manipulate and solve matrices is a fundamental skill in linear algebra and extends to more advanced topics like vector spaces and eigenvalues.

Substitution Method

The substitution method is a classic approach to solving systems of linear equations, where one equation is solved for one variable and then that solution is substituted into other equations. This method works elegantly when we can easily isolate a variable, as seen in our exercise with the third equation, where we found that \( z = -4 \).

Once we have a value for one variable, we substitute it into the other equations to find the remaining unknowns, reducing the system step by step. This approach is particularly helpful when dealing with smaller systems or when dealing with a system that naturally lends itself to isolating variables, as it can lead to a rapid solution without the need for more complex algebraic manipulations.

While powerful, the substitution method can become cumbersome with larger systems or when it's difficult to isolate variables. In such cases, students might choose matrix methods or the elimination method as alternatives.

Consistent Systems

When we talk about consistent systems in the context of linear equations, we mean that a system has at least one set of values for the variables that satisfies all the equations simultaneously. Consistency is crucial because it tells us that a solution exists and we can therefore search for it using methods like substitution or matrices.

In the given exercise, both systems ultimately yield the same solution set \( \{x=1, y=6, z=-4\} \), demonstrating that they are indeed consistent systems. Consistency can be identified in various ways: through graphing and observing if the lines or planes intersect, or through algebraic methods such as reducing the system to row-echelon form and checking for contradictions like '0 = 1'.

As students progress in their study of linear algebra, they'll encounter different types of consistent systems, such as those that are not only consistent but also independent (having a single unique solution) or dependent (having infinitely many solutions). Recognizing these differences is essential for solving and understanding the behavior of linear systems.

Linear Algebra

Linear algebra is a substantial branch of mathematics that deals with vectors, vector spaces, linear transformations, and systems of linear equations. It is foundational for understanding topics in engineering, physics, computer science, and more. In our exercise, we took advantage of linear algebra concepts by transforming systems of linear equations into matrix form to find their solutions.

The versatility of linear algebra stems from its ability to abstract problems so they can be analyzed and solved in a generalized way. The matrix methods and the substitution method we've discussed are tools within this field that enable the solution of not only simple equations but also more complex multidimensional problems.

Studying linear algebra opens doors to understanding spaces of different dimensions, calculating areas and volumes, and even working with data in fields such as machine learning. It encourages a structural approach to problem-solving, which is an invaluable skill set in many areas of science and technology.