Hey everyone! Today, we're diving into a powerful technique in linear algebra called Gauss-Jordan elimination. This method is a systematic way to solve systems of linear equations, and it's super useful in various fields like engineering, economics, and computer science. We'll walk through a step-by-step example to show you exactly how it works.
What is Gauss-Jordan Elimination?
Before we jump into our example, let's quickly define what Gauss-Jordan elimination is all about. At its core, this method uses a series of elementary row operations on an augmented matrix to transform the system of equations into a reduced row echelon form. Sounds complicated? Don't worry, we'll break it down. The key idea is to manipulate the equations (rows in our matrix) in a way that isolates each variable, making the solution crystal clear. Gauss-Jordan elimination is essentially a refined version of Gaussian elimination. While Gaussian elimination brings the matrix to row echelon form (upper triangular form), Gauss-Jordan goes a step further to achieve reduced row echelon form, where each leading entry (pivot) is 1, and all other entries in the same column are 0. This final form directly gives us the solution to the system of equations.
Think of it like this: you have a puzzle, and each equation is a piece. Gauss-Jordan elimination is the process of rearranging and manipulating those pieces until the solution (the complete picture) becomes obvious. The elementary row operations are the tools we use to move and adjust the pieces. These operations include:
- Swapping two rows.
- Multiplying a row by a non-zero constant.
- Adding a multiple of one row to another row.
These operations are crucial because they don't change the solution set of the system. They just rewrite the equations in a more convenient form. So, Gauss-Jordan elimination is our reliable tool for systematically solving linear equations. It provides a clear path to the solution, even for systems with many variables. Now, let's get into the nitty-gritty and solve a system of equations together. We'll take it slow and steady, so you can see every step along the way. Remember, practice makes perfect, so don't hesitate to try out more examples on your own.
Setting Up the Augmented Matrix
Okay, guys, let's get to the fun part! We're going to use Gauss-Jordan elimination to solve the following system of equations:
3x + 5y = 7
6x - y = -8
The first step in Gauss-Jordan elimination is to represent this system as an augmented matrix. An augmented matrix is just a compact way of writing the system, focusing on the coefficients and constants. We create the matrix by taking the coefficients of the variables (x and y) and the constants on the right side of the equations. So, for our system, the augmented matrix looks like this:
[ 3 5 | 7 ]
[ 6 -1 | -8 ]
See how we've arranged the numbers? The first row corresponds to the first equation (3x + 5y = 7), and the second row corresponds to the second equation (6x - y = -8). The vertical line separates the coefficient matrix (the part with the coefficients of x and y) from the constant vector (the part with the constants on the right side). This representation makes it easier to perform row operations without having to rewrite the variables and equals signs every time. It's a clean and efficient way to handle the system. The matrix is a powerful tool that allows us to manipulate the equations in a structured manner. Each row represents an equation, and each column represents the coefficients of a specific variable (or the constants). By performing elementary row operations on the matrix, we're essentially performing algebraic manipulations on the original equations. The beauty of the augmented matrix is that it keeps everything organized and allows us to focus on the numerical relationships within the system. It is very important to set up the augmented matrix correctly because any error here will propagate through the rest of the solution. Double-check that you've entered the coefficients and constants in the correct positions.
Performing Elementary Row Operations
Now that we have our augmented matrix, the real magic begins! We're going to perform a series of elementary row operations to transform the matrix into reduced row echelon form. Remember those operations we talked about earlier? Swapping rows, multiplying a row by a constant, and adding a multiple of one row to another – these are our tools. The goal is to get a matrix where the left side (the coefficient matrix) looks like the identity matrix:
[ 1 0 ]
[ 0 1 ]
This means we want a '1' in the top-left corner (the pivot), a '0' below it, a '1' in the bottom-right corner, and a '0' above it. The right side of the matrix will then give us the values of x and y directly. Let's start by getting a '1' in the top-left corner. We can do this by dividing the first row by 3:
[ 3 5 | 7 ] -> (1/3) * Row 1 -> [ 1 5/3 | 7/3 ]
[ 6 -1 | -8 ] [ 6 -1 | -8 ]
Now we have a '1' in the top-left! Next, we want a '0' below this '1'. To achieve this, we can subtract 6 times the first row from the second row:
[ 1 5/3 | 7/3 ] [ 1 5/3 | 7/3 ]
[ 6 -1 | -8 ] -> Row 2 - 6 * Row 1 -> [ 0 -11 | -22 ]
Great! We've got a '0' in the desired position. Now, let's focus on the second row. We want a '1' in the bottom-right corner. We can get this by dividing the second row by -11:
[ 1 5/3 | 7/3 ] [ 1 5/3 | 7/3 ]
[ 0 -11 | -22 ] -> (1/-11) * Row 2 -> [ 0 1 | 2 ]
Perfect! We're almost there. The last step is to get a '0' above the '1' in the second row. We can do this by subtracting (5/3) times the second row from the first row:
[ 1 5/3 | 7/3 ] -> Row 1 - (5/3) * Row 2 -> [ 1 0 | -1 ]
[ 0 1 | 2 ] [ 0 1 | 2 ]
And voilà! We've transformed our matrix into reduced row echelon form. You might be thinking, "Wow, that was a lot of steps!" And you're right, it can be. But with practice, you'll get faster and more comfortable with these operations. The key is to be methodical and keep track of what you're doing. Each step brings you closer to the solution. Remember, the goal is to systematically eliminate variables until you have a clear picture of the values of x and y. These elementary row operations are the heart of the Gauss-Jordan elimination process. They allow us to manipulate the equations in a way that preserves the solution while simplifying the system. By carefully choosing which operations to perform, we can systematically isolate each variable and arrive at the solution. So, keep practicing these operations, and you'll become a master of Gauss-Jordan elimination in no time!
Reading the Solution
Alright, we've done the hard work of transforming the augmented matrix into reduced row echelon form. Now comes the easy and rewarding part: reading off the solution! Remember, our final matrix looks like this:
[ 1 0 | -1 ]
[ 0 1 | 2 ]
This matrix directly tells us the values of x and y. The first row corresponds to the equation 1x + 0y = -1, which simplifies to x = -1. The second row corresponds to the equation 0x + 1y = 2, which simplifies to y = 2. So, the solution to our system of equations is x = -1 and y = 2! See how clean and straightforward that is? The reduced row echelon form makes the solution crystal clear. There's no need to substitute back into the original equations or do any further algebra. The solution is right there in the matrix! This is one of the great advantages of Gauss-Jordan elimination. It not only solves the system but also presents the solution in a readily understandable format. To be absolutely sure we have the correct solution, it's always a good idea to check our answer by plugging the values of x and y back into the original equations. Let's do that now:
3x + 5y = 7 -> 3(-1) + 5(2) = -3 + 10 = 7 (Correct!)
6x - y = -8 -> 6(-1) - 2 = -6 - 2 = -8 (Correct!)
Our solution checks out! This step is crucial to catch any errors we might have made along the way. It's a simple way to ensure that our answer is correct and that we've successfully solved the system. So, we've successfully used Gauss-Jordan elimination to solve the system of equations. We started by setting up the augmented matrix, performed elementary row operations to transform it into reduced row echelon form, and then simply read off the solution. This method is powerful because it provides a systematic approach to solving linear systems, no matter how many variables or equations there are. With practice, you'll become proficient in using Gauss-Jordan elimination and be able to tackle even more complex systems. Remember, the key is to understand the process, be methodical in your steps, and always check your answer!
Conclusion
Guys, we've reached the end of our journey through Gauss-Jordan elimination! We've seen how this method provides a systematic way to solve systems of linear equations by transforming the augmented matrix into reduced row echelon form. From setting up the matrix to performing elementary row operations and finally reading the solution, we've covered all the key steps. Gauss-Jordan elimination is a powerful tool in linear algebra, and mastering it will open doors to solving a wide range of problems in various fields. It is more than just a technique; it's a way of thinking about systems of equations in a structured and organized manner. By breaking down the problem into smaller steps and using elementary row operations, we can systematically eliminate variables and arrive at the solution. The reduced row echelon form is the final destination, where the solution is presented in a clear and concise way. One of the most important takeaways is the importance of being methodical and careful in each step. Errors can easily creep in if we're not paying attention, so it's crucial to double-check our work and ensure that we're performing the row operations correctly. Practice is key to mastering this technique. The more you work through examples, the more comfortable you'll become with the process, and the faster you'll be able to solve systems of equations. So, don't be afraid to tackle different problems and challenge yourself. Remember to always check your solution by plugging the values back into the original equations. This is a simple but effective way to ensure that you've arrived at the correct answer. Gauss-Jordan elimination is a fundamental concept in linear algebra, and its applications extend far beyond just solving systems of equations. It's used in matrix inversions, finding determinants, and solving linear programming problems, among other things. By understanding the principles behind Gauss-Jordan elimination, you'll gain a solid foundation for further studies in mathematics and related fields. So, keep practicing, keep exploring, and keep applying this powerful tool to new and exciting problems! Whether you're solving equations in engineering, economics, or computer science, Gauss-Jordan elimination will be a valuable asset in your problem-solving toolkit.