Solving Equations With Gauss-Jordan Elimination
Hey everyone! Today, we're diving deep into the world of linear equations and tackling a common problem: solving a system of equations. We'll be using a powerful technique called Gauss-Jordan elimination. Trust me, it sounds intimidating, but it's actually a systematic way to unravel these equations and find our solutions. So, grab your pencils, and let's get started!
Understanding Gauss-Jordan Elimination
Before we jump into the problem, let's quickly recap what Gauss-Jordan elimination is all about. In essence, it's a method to transform a system of linear equations into an equivalent, simpler system that's much easier to solve. We do this by manipulating the equations using elementary row operations. These operations are:
- Swapping two rows.
- Multiplying a row by a non-zero constant.
- Adding a multiple of one row to another row.
The ultimate goal is to get the system into what we call reduced row-echelon form. This form has a clear, diagonal pattern that directly tells us the values of our variables. Think of it like a secret code we're cracking, and reduced row-echelon form is the key to unlocking the solution.
The Gauss-Jordan elimination method is a systematic approach to solving systems of linear equations. It's a powerful tool because it works for systems of any size, whether you have two equations and two unknowns, or ten equations and ten unknowns! The beauty of the method lies in its organized nature. By following the steps carefully, you can methodically eliminate variables and isolate the solutions.
One of the key advantages of Gauss-Jordan elimination is that it can handle systems with unique solutions, infinitely many solutions, or no solutions at all. If the system has a unique solution, the reduced row-echelon form will clearly show the values of each variable. If there are infinitely many solutions, you'll see rows of zeros in the reduced form, indicating that some variables can be expressed in terms of others. And if there are no solutions, you'll encounter a contradiction in the reduced form, like an equation that says 0 = 1. This versatility makes Gauss-Jordan elimination a valuable technique in various fields, from mathematics and physics to engineering and economics. It's a fundamental tool for anyone working with systems of linear equations.
Let's Solve an Example
Okay, enough theory! Let's put Gauss-Jordan elimination into action. We're given the following system of equations:
x + y + z = 2
2x - 3y + z = -11
-x + 2y - z = 8
Our mission is to find the values of x, y, and z that satisfy all three equations simultaneously.
Step 1: Represent the System as an Augmented Matrix
First, we'll rewrite the system as an augmented matrix. This is just a compact way to represent the equations, focusing on the coefficients and constants. Our augmented matrix will look like this:
[ 1 1 1 | 2 ]
[ 2 -3 1 | -11 ]
[-1 2 -1 | 8 ]
The vertical line separates the coefficient matrix (the numbers in front of the variables) from the constant terms on the right side of the equations.
Step 2: Perform Row Operations to Get to Reduced Row-Echelon Form
Now comes the fun part! We'll use elementary row operations to transform this matrix into reduced row-echelon form. Remember, our goal is to get a matrix that looks like this:
[ 1 0 0 | x ]
[ 0 1 0 | y ]
[ 0 0 1 | z ]
Where x, y, and z will be our solutions.
Let's start by eliminating the 2 in the first column of the second row. We can do this by multiplying the first row by -2 and adding it to the second row:
R2 = R2 - 2 * R1
This gives us:
[ 1 1 1 | 2 ]
[ 0 -5 -1 | -15 ]
[-1 2 -1 | 8 ]
Next, let's eliminate the -1 in the first column of the third row. We can do this by adding the first row to the third row:
R3 = R3 + R1
This gives us:
[ 1 1 1 | 2 ]
[ 0 -5 -1 | -15 ]
[ 0 3 0 | 10 ]
Now, let's get a 1 in the second row, second column. We can do this by dividing the second row by -5:
R2 = R2 / -5
This gives us:
[ 1 1 1 | 2 ]
[ 0 1 1/5 | 3 ]
[ 0 3 0 | 10 ]
Next, let's eliminate the 3 in the second column of the third row. We can do this by multiplying the second row by -3 and adding it to the third row:
R3 = R3 - 3 * R2
This gives us:
[ 1 1 1 | 2 ]
[ 0 1 1/5 | 3 ]
[ 0 0 -3/5 | 1 ]
Now, let's get a 1 in the third row, third column. We can do this by multiplying the third row by -5/3:
R3 = R3 * -5/3
This gives us:
[ 1 1 1 | 2 ]
[ 0 1 1/5 | 3 ]
[ 0 0 1 | -5/3 ]
We're getting closer! Now, we need to eliminate the 1/5 in the second row, third column. We can do this by multiplying the third row by -1/5 and adding it to the second row:
R2 = R2 - 1/5 * R3
This gives us:
[ 1 1 1 | 2 ]
[ 0 1 0 | 10/3 ]
[ 0 0 1 | -5/3 ]
Next, let's eliminate the 1 in the first row, third column. We can do this by subtracting the third row from the first row:
R1 = R1 - R3
This gives us:
[ 1 1 0 | 11/3 ]
[ 0 1 0 | 10/3 ]
[ 0 0 1 | -5/3 ]
Finally, let's eliminate the 1 in the first row, second column. We can do this by subtracting the second row from the first row:
R1 = R1 - R2
This gives us:
[ 1 0 0 | 1/3 ]
[ 0 1 0 | 10/3 ]
[ 0 0 1 | -5/3 ]
Step 3: Interpret the Solution
We've reached reduced row-echelon form! Now we can directly read off the solution:
x = 1/3
y = 10/3
z = -5/3
So, the solution to the system of equations is x = 1/3, y = 10/3, and z = -5/3.
Analyzing the Given Options
Now, let's take a look at the options provided in the original question:
A. x = 1, y = 3, z = 2 B. x = 3, y = -1, z = 0
Neither of these options matches the solution we found using Gauss-Jordan elimination. This means that neither option A nor option B is the correct answer. It's possible that there was a mistake in the original options provided, or perhaps there are more options that were not included in the prompt.
Key Takeaways
- Gauss-Jordan elimination is a powerful method for solving systems of linear equations.
- The process involves transforming the system into an augmented matrix and then using elementary row operations to reach reduced row-echelon form.
- The reduced row-echelon form directly reveals the solution to the system.
- Gauss-Jordan elimination can handle systems with unique solutions, infinitely many solutions, or no solutions.
Practice Makes Perfect
The best way to master Gauss-Jordan elimination is to practice! Try solving different systems of equations using this method. You'll become more comfortable with the row operations and the overall process. And remember, if you get stuck, there are plenty of resources available online and in textbooks to help you out.
So, there you have it! We've successfully used Gauss-Jordan elimination to solve a system of equations. Keep practicing, and you'll be solving linear systems like a pro in no time!