Mastering 3x3 Systems: Gaussian, Gauss-Jordan, Cramer's Rule
Unlocking the Secrets of Linear Systems
Hey there, math enthusiasts and problem-solvers! Ever found yourself staring at a set of equations, wondering how to untangle them? Well, you're in the right place! Linear systems are super important in mathematics, engineering, economics, and pretty much everywhere you look. They help us model real-world scenarios, from predicting market trends to designing complex structures. Today, we're going to dive deep into solving a specific 3x3 system of linear equations using not one, not two, but three powerful methods: Gaussian Elimination, Gauss-Jordan Elimination, and Cramer's Rule. Each method offers a unique approach and its own set of advantages, making them valuable tools in your mathematical arsenal. Whether you're a student trying to ace your next exam or just someone who loves the thrill of solving a good puzzle, understanding these techniques will seriously level up your problem-solving game. We'll break down each method step-by-step, making sure everything is super clear and easy to follow. So, grab a coffee, get comfortable, and let's embark on this exciting journey to master linear systems!
Seriously, guys, learning these methods isn't just about getting the right answer; it's about understanding the underlying logic and appreciating the elegance of mathematics. Each technique we'll explore today showcases a different facet of linear algebra, a field that's all about understanding relationships and transformations. We'll start with Gaussian Elimination, which is like the fundamental building block for many numerical methods, gradually simplifying our system until it's a breeze to solve. Then, we'll crank it up a notch with Gauss-Jordan, taking that simplification even further to get our answers practically handed to us. And finally, for those who love determinants and a more formulaic approach, Cramer's Rule will be our go-to. Don't worry if these names sound intimidating right now; by the end of this article, you'll be speaking the language of linear algebra like a pro. Our goal isn't just to solve this particular problem, but to equip you with the knowledge and confidence to tackle any similar system you might encounter. Ready to transform those complex equations into simple solutions? Let's get started!
The System We're Tackling
Before we dive into the nitty-gritty of the solutions, let's clearly lay out the specific 3x3 system of linear equations we'll be working with throughout this article. This is our target, the puzzle we aim to solve using all three methods. Having it clearly in sight will help us appreciate how each method systematically leads us to the same correct answer.
Here's the system:
- 5x₁ + 2x₂ - 2x₃ = 1
- x₁ + 5x₂ - 3x₃ = -2
- 5x₁ - 3x₂ + 5x₃ = 2
Our mission, should we choose to accept it (and we definitely do!), is to find the unique values for x₁, x₂, and x₃ that satisfy all three of these equations simultaneously. Let's get to it!
Method 1: Unraveling Systems with Gaussian Elimination
Alright, folks, let's kick things off with Gaussian Elimination, a cornerstone method for solving linear systems. This technique is all about transforming our system into an upper triangular matrix through a series of elementary row operations. Think of it like a systematic way to simplify the equations step-by-step until they become super easy to solve using back-substitution. The main keywords here are row operations, echelon form, and back-substitution. We're essentially trying to create zeros below the main diagonal, isolating our variables one by one. This process is incredibly powerful and forms the basis for many computational algorithms. It's often the first method taught for its logical, step-by-step nature, guiding us from a complex system to a straightforward solution. The beauty of Gaussian elimination lies in its ability to handle systems of any size, not just 3x3, making it a truly versatile tool in linear algebra. We'll start by representing our system as an augmented matrix, which is simply a compact way to write down the coefficients and constants of our equations. This matrix representation makes performing row operations much cleaner and less prone to errors than juggling multiple equations simultaneously. Trust me, once you get the hang of it, you'll find it quite intuitive!
Let's apply Gaussian Elimination to our system:
5x₁ + 2x₂ - 2x₃ = 1 x₁ + 5x₂ - 3x₃ = -2 5x₁ - 3x₂ + 5x₃ = 2
First, we write this as an augmented matrix:
[ 5 2 -2 | 1 ]
[ 1 5 -3 | -2 ]
[ 5 -3 5 | 2 ]
Our goal is to get this into row echelon form. This means getting ones on the main diagonal and zeros below them. Let's make the entry in the top-left corner a '1' to start, which is often called the pivot element. We can achieve this by swapping Row 1 and Row 2.
Operation: R₁ ↔ R₂
[ 1 5 -3 | -2 ]
[ 5 2 -2 | 1 ]
[ 5 -3 5 | 2 ]
Now, we want to create zeros below this leading '1' in the first column. This involves subtracting multiples of the first row from the rows below it.
Operation: R₂ ← R₂ - 5R₁ Operation: R₃ ← R₃ - 5R₁
[ 1 5 -3 | -2 ]
[ 0 -23 13 | 11 ] (2 - 5*5 = -23; -2 - 5*(-3) = 13; 1 - 5*(-2) = 11)
[ 0 -28 20 | 12 ] (-3 - 5*5 = -28; 5 - 5*(-3) = 20; 2 - 5*(-2) = 12)
Next, we focus on the second column, aiming for a '1' in the second row, second column position (our new pivot) and a '0' below it. To simplify our numbers a bit, we can divide the third row by 4.
Operation: R₃ ← (1/4)R₃
[ 1 5 -3 | -2 ]
[ 0 -23 13 | 11 ]
[ 0 -7 5 | 3 ] (Divide -28, 20, 12 by 4)
Now, let's get a '1' in the R₂C₂ position. Dividing R₂ by -23 is an option, but it introduces fractions early. Alternatively, we can use a combination to make the R₃C₂ entry zero first. Let's divide R₂ by -23 to make the pivot 1. This is a standard step in Gaussian elimination.
Operation: R₂ ← (-1/23)R₂
[ 1 5 -3 | -2 ]
[ 0 1 -13/23 | -11/23 ] (13 / -23 = -13/23; 11 / -23 = -11/23)
[ 0 -7 5 | 3 ]
Now, create a zero below the leading '1' in the second column.
Operation: R₃ ← R₃ + 7R₂
[ 1 5 -3 | -2 ]
[ 0 1 -13/23 | -11/23 ]
[ 0 0 (5 + 7*(-13/23)) | (3 + 7*(-11/23)) ]
Let's calculate those fractions: 5 + 7*(-13/23) = 5 - 91/23 = (115 - 91)/23 = 24/23. And 3 + 7*(-11/23) = 3 - 77/23 = (69 - 77)/23 = -8/23.
So, our matrix in row echelon form is:
[ 1 5 -3 | -2 ]
[ 0 1 -13/23 | -11/23 ]
[ 0 0 24/23 | -8/23 ]
From this form, we can use back-substitution to find our variables. The last row represents the equation: (24/23)x₃ = -8/23. Multiplying both sides by 23, we get 24x₃ = -8, so x₃ = -8/24 = -1/3.
Now substitute x₃ = -1/3 into the second row's equation: x₂ - (13/23)x₃ = -11/23. x₂ - (13/23)(-1/3) = -11/23 x₂ + 13/69 = -11/23 x₂ = -11/23 - 13/69 = (-33 - 13)/69 = -46/69 = -2/3.
Finally, substitute x₃ = -1/3 and x₂ = -2/3 into the first row's equation: x₁ + 5x₂ - 3x₃ = -2. x₁ + 5(-2/3) - 3(-1/3) = -2 x₁ - 10/3 + 1 = -2 x₁ - 7/3 = -2 x₁ = -2 + 7/3 = (-6 + 7)/3 = 1/3.
So, the solution to our system using Gaussian Elimination is x₁ = 1/3, x₂ = -2/3, x₃ = -1/3. Pretty neat, right? This method systematically brought us to the solution by simplifying the system into a more manageable form.
Method 2: Going Further with Gauss-Jordan Elimination
Building on the foundations of Gaussian Elimination, we now turn our attention to Gauss-Jordan Elimination. If Gaussian Elimination aims for row echelon form, Gauss-Jordan takes it a step further to achieve reduced row echelon form. What's the difference, you ask? Well, in reduced row echelon form, not only do we have leading '1's in each non-zero row and zeros below them, but we also have zeros above them. This means that instead of using back-substitution, the solution literally appears in the augmented column of the matrix, creating what looks like an identity matrix on the left side. It's like Gaussian Elimination gets you 90% of the way there, and Gauss-Jordan finishes the job to give you the answer directly. This method is incredibly elegant because once the matrix is transformed, you can simply read off the values of your variables without any further calculation. For this reason, it's often preferred in computational settings where direct answers are key. Understanding this method really highlights the power of matrix transformations and how they can simplify complex problems into straightforward readings. The main keywords here are reduced row echelon form, identity matrix, and direct solution extraction. Let's see how we can apply this powerful technique to our system, continuing from our row echelon form from the Gaussian elimination, or starting fresh to illustrate the full process.
To ensure we cover the complete Gauss-Jordan Elimination process from the start for clarity, let's revisit our initial augmented matrix:
[ 5 2 -2 | 1 ]
[ 1 5 -3 | -2 ]
[ 5 -3 5 | 2 ]
Just like with Gaussian Elimination, we'll start by swapping R₁ and R₂ to get a leading '1' in the first row, first column:
Operation: R₁ ↔ R₂
[ 1 5 -3 | -2 ]
[ 5 2 -2 | 1 ]
[ 5 -3 5 | 2 ]
Next, we create zeros below the leading '1' in the first column:
Operation: R₂ ← R₂ - 5R₁ Operation: R₃ ← R₃ - 5R₁
[ 1 5 -3 | -2 ]
[ 0 -23 13 | 11 ]
[ 0 -28 20 | 12 ]
We can simplify R₃ by dividing by 4:
Operation: R₃ ← (1/4)R₃
[ 1 5 -3 | -2 ]
[ 0 -23 13 | 11 ]
[ 0 -7 5 | 3 ]
Now, we aim for a '1' in the second row, second column. Divide R₂ by -23:
Operation: R₂ ← (-1/23)R₂
[ 1 5 -3 | -2 ]
[ 0 1 -13/23 | -11/23 ]
[ 0 -7 5 | 3 ]
Create a zero below this new leading '1' in the second column:
Operation: R₃ ← R₃ + 7R₂
[ 1 5 -3 | -2 ]
[ 0 1 -13/23 | -11/23 ]
[ 0 0 24/23 | -8/23 ]
We are now in row echelon form, which is where Gaussian Elimination stops. For Gauss-Jordan, we continue! We need to make the last leading entry a '1'.
Operation: R₃ ← (23/24)R₃
[ 1 5 -3 | -2 ]
[ 0 1 -13/23 | -11/23 ]
[ 0 0 1 | -8/24 = -1/3 ]
Now that we have leading '1's on the diagonal, the next step in Gauss-Jordan is to create zeros above these leading '1's, starting from the last column and working our way up. We'll start by making the entries in the third column above the '1' in R₃C₃ into zeros.
Operation: R₂ ← R₂ + (13/23)R₃ Operation: R₁ ← R₁ + 3R₃
Let's calculate R₂ first: R₂C₄: -11/23 + (13/23)*(-1/3) = -11/23 - 13/69 = (-33 - 13)/69 = -46/69 = -2/3
Now calculate R₁: R₁C₄: -2 + 3*(-1/3) = -2 - 1 = -3. Uh oh, something's not right. Let's recheck the calculation of R1C4 carefully.
Okay, let's re-calculate R₁ ← R₁ + 3R₃: For the constant term: -2 + 3 * (-1/3) = -2 - 1 = -3. This seems correct for the first constant, but our expected solution for x1 is 1/3. This means there's a problem in my mental re-calculation, or previous steps. Let me re-verify the matrix before these final steps.
Previous matrix:
[ 1 5 -3 | -2 ]
[ 0 1 -13/23 | -11/23 ]
[ 0 0 1 | -1/3 ]
Operation: R₂ ← R₂ + (13/23)R₃
- R₂C₃: -13/23 + (13/23)*1 = 0 (Good)
- R₂C₄: -11/23 + (13/23)*(-1/3) = -11/23 - 13/69 = (-33 - 13)/69 = -46/69 = -2/3 (Good)
New R₂ is now: [ 0 1 0 | -2/3 ]
Operation: R₁ ← R₁ + 3R₃
- R₁C₃: -3 + 3*1 = 0 (Good)
- R₁C₄: -2 + 3*(-1/3) = -2 - 1 = -3. Still getting -3 here. This means R1C2 needs to be zeroed out after R1C3 is zeroed out.
Ah, I see my mistake! When performing Gauss-Jordan, we should proceed column by column. We zeroed out elements below pivots, then make the pivots 1. Now, we zero out elements above the pivots, starting from the last column's pivot and moving left. So, it should be:
Matrix is:
[ 1 5 -3 | -2 ]
[ 0 1 -13/23 | -11/23 ]
[ 0 0 1 | -1/3 ]
-
Make R₂C₃ zero: R₂ ← R₂ + (13/23)R₃
[ 1 5 -3 | -2 ] [ 0 1 0 | -2/3 ] (Calculated: -11/23 + (13/23)(-1/3) = -46/69 = -2/3) [ 0 0 1 | -1/3 ] -
Make R₁C₃ zero: R₁ ← R₁ + 3R₃
[ 1 5 0 | -3 ] (Calculated: -2 + 3(-1/3) = -2 - 1 = -3) [ 0 1 0 | -2/3 ] [ 0 0 1 | -1/3 ]
Now, we have a leading '1' in R₂C₂. We need to make R₁C₂ zero. This is the last step for reduced row echelon form.
Operation: R₁ ← R₁ - 5R₂
- R₁C₄: -3 - 5*(-2/3) = -3 + 10/3 = (-9 + 10)/3 = 1/3.
Finally, our matrix is in reduced row echelon form:
[ 1 0 0 | 1/3 ]
[ 0 1 0 | -2/3 ]
[ 0 0 1 | -1/3 ]
And there you have it! The solution can be read directly from the last column. x₁ = 1/3, x₂ = -2/3, x₃ = -1/3. This matches the solution we found with Gaussian Elimination, but without the need for back-substitution. Gauss-Jordan is a bit more work up front in terms of row operations, but the payoff is a direct answer at the end, which is incredibly satisfying and efficient for certain applications!
Method 3: Decoding with Cramer's Rule
Now for something completely different, yet equally powerful: Cramer's Rule! This method offers an alternative approach to solving systems of linear equations, primarily leveraging the concept of determinants. If you're someone who loves formulas and a direct calculation, Cramer's Rule might just become your new favorite. It's particularly elegant for smaller systems, like our 3x3, as it provides a clear formula for each variable. However, it can get computationally intensive for much larger systems due to the number of determinants that need to be calculated. The main keywords for this section are determinants, coefficient matrix, and replacing columns. We'll learn how to calculate the determinant of the original coefficient matrix and then use it in conjunction with determinants of modified matrices to find each variable individually. The beauty of Cramer's Rule is that it gives us a direct, explicit formula for each unknown, which can be super handy. It also provides insight into when a unique solution exists (i.e., when the main determinant is non-zero).
Let's apply Cramer's Rule to our system:
5x₁ + 2x₂ - 2x₃ = 1 x₁ + 5x₂ - 3x₃ = -2 5x₁ - 3x₂ + 5x₃ = 2
First, we need to extract the coefficient matrix (A) and the constant vector (B):
A = [ 5 2 -2 ] _ [ 1 5 -3 ]_ _ [ 5 -3 5 ]_
B = [ 1 ] _ [-2 ]_ _ [ 2 ]_
Step 1: Calculate the determinant of the coefficient matrix, D (or det(A)). For a 3x3 matrix, we can use the Sarrus' Rule or cofactor expansion. Let's use Sarrus' rule for clarity.
D = 5(55 - (-3)(-3)) - 2(15 - (-3)5) + (-2)(1(-3) - 55) D = 5(25 - 9) - 2(5 + 15) - 2(-3 - 25) D = 5(16) - 2(20) - 2(-28) D = 80 - 40 + 56 D = 40 + 56 = 96
Since D ≠ 0, a unique solution exists, which is great news!
Step 2: Calculate the determinant Dx₁. To do this, we replace the first column of the coefficient matrix A with the constant vector B.
Dx₁ = [ 1 2 -2 ] _ [-2 5 -3 ]_ _ [ 2 -3 5 ]_
Dx₁ = 1(55 - (-3)(-3)) - 2((-2)5 - (-3)2) + (-2)((-2)(-3) - 52) Dx₁ = 1(25 - 9) - 2(-10 + 6) - 2(6 - 10) Dx₁ = 1(16) - 2(-4) - 2(-4) Dx₁ = 16 + 8 + 8 = 32
Step 3: Calculate the determinant Dx₂. Here, we replace the second column of the coefficient matrix A with the constant vector B.
Dx₂ = [ 5 1 -2 ] _ [ 1 -2 -3 ]_ _ [ 5 2 5 ]_
Dx₂ = 5((-2)*5 - (-3)2) - 1(15 - (-3)5) + (-2)(12 - (-2)*5) Dx₂ = 5(-10 + 6) - 1(5 + 15) - 2(2 + 10) Dx₂ = 5(-4) - 1(20) - 2(12) Dx₂ = -20 - 20 - 24 = -64
Step 4: Calculate the determinant Dx₃. For this, we replace the third column of the coefficient matrix A with the constant vector B.
Dx₃ = [ 5 2 1 ] _ [ 1 5 -2 ]_ _ [ 5 -3 2 ]_
Dx₃ = 5(52 - (-2)(-3)) - 2(12 - (-2)5) + 1(1(-3) - 55) Dx₃ = 5(10 - 6) - 2(2 + 10) + 1(-3 - 25) Dx₃ = 5(4) - 2(12) + 1(-28) Dx₃ = 20 - 24 - 28 Dx₃ = -4 - 28 = -32
Step 5: Use Cramer's Rule formula to find the values of x₁, x₂, x₃.
x₁ = Dx₁ / D = 32 / 96 = 1/3 x₂ = Dx₂ / D = -64 / 96 = -2/3 x₃ = Dx₃ / D = -32 / 96 = -1/3
And there we have it! Once again, the solution is x₁ = 1/3, x₂ = -2/3, x₃ = -1/3. Cramer's Rule provides a very structured way to find each variable, requiring a bit of determinant calculation but offering a direct path to the answer without row operations or back-substitution once the determinants are found. It's a fantastic example of how different mathematical concepts (matrices and determinants) come together to solve a common problem!
Which Method Should You Use?
Okay, so we've just journeyed through three distinct and incredibly useful methods for solving systems of linear equations: Gaussian Elimination, Gauss-Jordan Elimination, and Cramer's Rule. You might be wondering,