Solve 3x3 Equations With Gauss Elimination: Step-by-Step
Hey guys! Today, we're diving into a super important topic in math: solving systems of linear equations. Specifically, we're going to tackle a 3x3 system using a method called Gauss elimination. Trust me, this method is a game-changer once you get the hang of it. We'll break it down step by step, so don't worry if it looks intimidating at first.
What is Gauss Elimination?
So, what exactly is Gauss elimination? In a nutshell, it's a systematic way to solve systems of linear equations by transforming them into an upper triangular form. Think of it like climbing a staircase – we're eliminating variables one by one until we can easily solve for the remaining ones and then work our way back up. The method relies on performing elementary row operations, which are operations that don't change the solution of the system. These operations include:
- Swapping two rows: Imagine you have two equations, and you decide to switch their positions. This doesn't change the solution, right? That's the idea behind this operation.
- Multiplying a row by a non-zero constant: If you multiply both sides of an equation by the same number (other than zero), the equation is still balanced. We use this to get leading coefficients of 1.
- Adding a multiple of one row to another: This is the key operation for eliminating variables. By adding a multiple of one equation to another, we can strategically eliminate a variable from one of the equations.
Why use Gauss elimination? Well, it's a very organized and efficient method, especially for larger systems of equations. It helps to avoid messy algebra and reduces the chances of making mistakes. Plus, it's a fundamental technique that forms the basis for many other numerical methods.
To fully grasp the power of Gauss elimination, think of it as a powerful problem-solving tool that goes beyond just this specific example. It's a foundational technique in linear algebra, a branch of mathematics with countless applications in fields like engineering, computer science, economics, and statistics. Mastering Gauss elimination will not only help you ace your math exams but also equip you with a valuable skill for tackling real-world problems.
Setting Up the Problem
Okay, let's jump into our example problem. We've got this system of three equations with three unknowns (x, y, and z):
- x + 2y + z = 7
- 3x + y + z = 5
- 2x + 3y - z = 3
Our mission is to find the values of x, y, and z that satisfy all three equations simultaneously. The first step in Gauss elimination is to represent this system in an augmented matrix form. An augmented matrix is basically a shorthand notation that makes the calculations cleaner and easier to follow. It's like writing the coefficients of the variables and the constants in a table.
To create the augmented matrix, we'll take the coefficients of x, y, and z from each equation and arrange them in rows. The constants on the right-hand side of the equations will form the last column, separated by a vertical line (which represents the equals sign). So, our augmented matrix looks like this:
[ 1 2 1 | 7 ]
[ 3 1 1 | 5 ]
[ 2 3 -1 | 3 ]
Each row in the matrix represents one of the equations, and each column (except the last one) corresponds to a variable. The last column represents the constants. Now that we have our augmented matrix, we're ready to start the elimination process! The goal is to transform this matrix into an upper triangular form, which means we want to get zeros below the main diagonal (the diagonal from the top-left corner to the bottom-right corner).
Before we move on, let's pause and appreciate the beauty of this representation. By converting the system of equations into an augmented matrix, we've essentially distilled the problem down to its essence. We've stripped away the variable names and symbols, focusing purely on the numerical relationships between the equations. This not only simplifies the calculations but also reveals the underlying structure of the system, making it easier to manipulate and solve.
Step-by-Step Gauss Elimination
Alright, let's get our hands dirty and start the actual Gauss elimination! Remember, our goal is to get zeros below the main diagonal. We'll do this by performing elementary row operations.
Step 1: Eliminate x from the second and third equations
- To eliminate x from the second equation, we'll subtract 3 times the first row from the second row (R2 = R2 - 3R1). This will make the coefficient of x in the second row zero.
- To eliminate x from the third equation, we'll subtract 2 times the first row from the third row (R3 = R3 - 2R1). This will make the coefficient of x in the third row zero.
Let's perform these operations. First, R2 = R2 - 3R1:
[ 1 2 1 | 7 ]
[ 3-3(1) 1-3(2) 1-3(1) | 5-3(7) ] => [ 0 -5 -2 | -16 ]
[ 2 3 -1 | 3 ]
Next, R3 = R3 - 2R1:
[ 1 2 1 | 7 ]
[ 0 -5 -2 | -16 ]
[ 2-2(1) 3-2(2) -1-2(1) | 3-2(7) ] => [ 0 -1 -3 | -11 ]
Our matrix now looks like this:
[ 1 2 1 | 7 ]
[ 0 -5 -2 | -16 ]
[ 0 -1 -3 | -11 ]
See how we've got zeros in the first column below the diagonal? We're making progress! Now, let's move on to the next step.
Step 2: Eliminate y from the third equation
We want to eliminate y from the third equation. To do this, we'll subtract 1/5 times the second row from the third row (R3 = R3 - (1/5)R2). This will make the coefficient of y in the third row zero.
Let's perform this operation:
[ 1 2 1 | 7 ]
[ 0 -5 -2 | -16 ]
[ 0-(-1/5)(-5) -1-(-1/5)(-2) -3-(-1/5)(-2) | -11-(-1/5)(-16) ] => [ 0 0 -13/5 | -39/5 ]
Our matrix now looks like this:
[ 1 2 1 | 7 ]
[ 0 -5 -2 | -16 ]
[ 0 0 -13/5 | -39/5 ]
Step 3: Simplify the matrix (Optional but Recommended)
To make our lives easier, we can multiply the second row by -1/5 and the third row by -5/13 to get leading coefficients of 1:
- R2 = (-1/5)R2
- R3 = (-5/13)R3
Performing these operations, we get:
[ 1 2 1 | 7 ]
[ 0 1 2/5 | 16/5 ]
[ 0 0 1 | 3 ]
We've reached the upper triangular form! This is a fantastic milestone. Notice the staircase pattern of zeros below the main diagonal. Now, we can easily solve for the variables using back-substitution.
Back-Substitution: Finding the Solutions
Now comes the fun part: back-substitution! We've transformed our system into an upper triangular form, which makes solving for the variables a breeze. We'll start with the last equation and work our way up.
Step 1: Solve for z
The last row of our matrix represents the equation:
0x + 0y + 1z = 3
This simplifies to z = 3. Awesome! We've found our first variable.
Step 2: Solve for y
The second row of our matrix represents the equation:
0x + 1y + (2/5)z = 16/5
We already know that z = 3, so we can substitute that in:
y + (2/5)(3) = 16/5 y + 6/5 = 16/5 y = 16/5 - 6/5 y = 10/5 y = 2
Great! We've found y as well.
Step 3: Solve for x
The first row of our matrix represents the equation:
1x + 2y + 1z = 7
We know that y = 2 and z = 3, so we can substitute those in:
x + 2(2) + 3 = 7 x + 4 + 3 = 7 x + 7 = 7 x = 0
And there we have it! We've found x.
So, our solution is:
- x = 0
- y = 2
- z = 3
We can write this as an ordered triple: (0, 2, 3). This means that the point (0, 2, 3) is the intersection of the three planes represented by our original equations.
Before we celebrate too much, it's always a good idea to check our solution. We can do this by plugging the values of x, y, and z back into our original equations and making sure they hold true. Let's do it:
- Equation 1: 0 + 2(2) + 3 = 7 => 0 + 4 + 3 = 7 => 7 = 7 (Correct!)
- Equation 2: 3(0) + 2 + 3 = 5 => 0 + 2 + 3 = 5 => 5 = 5 (Correct!)
- Equation 3: 2(0) + 3(2) - 3 = 3 => 0 + 6 - 3 = 3 => 3 = 3 (Correct!)
Our solution checks out! We've successfully solved the system of equations using Gauss elimination and back-substitution.
Conclusion: Mastering Gauss Elimination
Woohoo! We did it! We successfully solved a 3x3 system of equations using Gauss elimination. I know it might seem like a lot of steps, but with practice, you'll become a pro at this. The key is to understand the underlying principles and to be organized with your calculations.
Gauss elimination is a powerful tool for solving systems of linear equations, and it's a fundamental concept in linear algebra. It's used in many different fields, from engineering and computer science to economics and finance. By mastering this technique, you're not just learning how to solve equations; you're developing a valuable problem-solving skill that will serve you well in many areas of life.
Remember, practice makes perfect! Try solving more 3x3 systems (and even larger ones!) using Gauss elimination. The more you practice, the more comfortable you'll become with the method. And don't be afraid to make mistakes – they're part of the learning process. Just learn from them, and keep going! You've got this!
So, next time you encounter a system of linear equations, remember the power of Gauss elimination. It's your secret weapon for conquering those equations and unlocking the solutions within. Keep practicing, keep exploring, and keep learning! You're on your way to becoming a math whiz!
