Gaussian Elimination: Solve System Of Equations

Hey guys! Today, we're diving into a fundamental technique in linear algebra: Gaussian elimination. This powerful method allows us to solve systems of linear equations, like the one we have here, or determine if no solution exists. We'll walk through the process step-by-step, making it super clear and easy to follow. So, let's get started and see how it works!

Understanding Gaussian Elimination

At its heart, Gaussian elimination is a systematic approach to transform a system of linear equations into an equivalent system that's much easier to solve. This transformation involves manipulating the equations using elementary row operations until we reach a form called row-echelon form or reduced row-echelon form. These forms make it straightforward to read off the solutions for the variables.

But what exactly are these elementary row operations? There are three main types:

1. Swapping two rows: This simply means interchanging the positions of two equations in the system. It's like reordering the list, which doesn't change the solution.
2. Multiplying a row by a nonzero constant: This involves multiplying both sides of an equation by the same number (except zero). It's like scaling an equation up or down, again without affecting the solution.
3. Adding a multiple of one row to another: This is where things get interesting. We can multiply one equation by a constant and then add it to another equation. This operation is crucial for eliminating variables and creating the desired echelon form.

The goal of these operations is to systematically eliminate variables from the equations until we have a triangular system. In this triangular system, the first equation will have all the variables, the second equation will have one less variable, the third equation will have two fewer variables, and so on. This makes it easy to solve for the variables one by one using back-substitution.

Before we jump into solving our specific system, it's helpful to represent the equations in matrix form. This makes the process more organized and visually clear. The matrix we'll be working with is called the augmented matrix, which combines the coefficients of the variables and the constants on the right-hand side of the equations.

So, why is Gaussian elimination so important? Well, it's not just a method for solving equations. It's a fundamental tool in various fields, including engineering, computer science, economics, and more. It's used to solve problems involving networks, circuits, optimization, and many other applications. Mastering this technique is a key step in understanding linear algebra and its applications.

Now, let's apply Gaussian elimination to our system of equations and see how it works in practice!

Setting Up the Augmented Matrix

Okay, let's get practical and apply Gaussian elimination to our system of equations. The first step is to represent the system in matrix form. This makes it easier to perform the row operations and keep track of our progress. We'll create what's called an augmented matrix, which combines the coefficients of the variables and the constants on the right-hand side of the equations.

Our system of equations is:

$$\left\{ \begin{array}{r} 2 x+9 y-8 z=1 \\ 3 x+12 y-10 z=5 \\ x+3 y-2 z=4 \end{array} \right.$$

To form the augmented matrix, we take the coefficients of x, y, and z from each equation and place them in the matrix. We also include the constants on the right-hand side of the equations, separated by a vertical line (which is optional but helps to keep things organized). So, the augmented matrix looks like this:

$$\left[ \begin{array}{rrr|r} 2 & 9 & -8 & 1 \\ 3 & 12 & -10 & 5 \\ 1 & 3 & -2 & 4 \end{array} \right]$$

Each row in the matrix represents one of our equations. The first three columns represent the coefficients of x, y, and z, respectively, and the last column represents the constants. This matrix representation allows us to perform row operations in a systematic way, without having to write out the variables every time. It's a much cleaner and more efficient way to work with the system of equations.

Now that we have our augmented matrix, the next step is to use elementary row operations to transform it into row-echelon form. Remember those three operations we talked about earlier? We'll be using them to manipulate the rows of the matrix until we get a triangular form, where the entries below the main diagonal are all zeros. This will make it easy to solve for the variables using back-substitution.

So, with our matrix ready to go, let's dive into the row operations and see how we can transform this matrix into a form that will reveal the solution to our system of equations. It might seem a bit like a puzzle at first, but once you get the hang of it, Gaussian elimination becomes a powerful tool in your mathematical arsenal.

Applying Gaussian Elimination Steps

Alright, guys, with our augmented matrix set up, let's roll up our sleeves and get into the heart of Gaussian elimination: applying those elementary row operations. Our goal here is to transform the matrix into row-echelon form, which means getting zeros below the main diagonal. This will make it super easy to solve the system using back-substitution.

Here’s our augmented matrix again to refresh our memory:

$$\left[ \begin{array}{rrr|r} 2 & 9 & -8 & 1 \\ 3 & 12 & -10 & 5 \\ 1 & 3 & -2 & 4 \end{array} \right]$$

Step 1: Get a 1 in the top-left corner

Ideally, we want a 1 in the top-left position (the first entry of the first row). We can achieve this by swapping the first and third rows. This is an elementary row operation, and it's perfectly valid:

$$R_1 \leftrightarrow R_3: \quad \left[ \begin{array}{rrr|r} 1 & 3 & -2 & 4 \\ 3 & 12 & -10 & 5 \\ 2 & 9 & -8 & 1 \end{array} \right]$$

Now we have a 1 in the top-left corner, which is exactly what we wanted!

Step 2: Get zeros below the leading 1 in the first column

Next, we want to get zeros below this leading 1. To do this, we'll use the third elementary row operation: adding a multiple of one row to another. We'll subtract 3 times the first row from the second row, and 2 times the first row from the third row:

$$R_2 \rightarrow R_2 - 3R_1: \quad \left[ \begin{array}{rrr|r} 1 & 3 & -2 & 4 \\ 0 & 3 & -4 & -7 \\ 2 & 9 & -8 & 1 \end{array} \right]$$

$$R_3 \rightarrow R_3 - 2R_1: \quad \left[ \begin{array}{rrr|r} 1 & 3 & -2 & 4 \\ 0 & 3 & -4 & -7 \\ 0 & 3 & -4 & -7 \end{array} \right]$$

Look at that! We now have zeros in the first column below the leading 1. We're making great progress!

Step 3: Get a leading 1 in the second row, second column

Now, let's focus on the second row. We want a 1 in the second column. We can achieve this by dividing the second row by 3:

$$R_2 \rightarrow \frac{1}{3}R_2: \quad \left[ \begin{array}{rrr|r} 1 & 3 & -2 & 4 \\ 0 & 1 & -\frac{4}{3} & -\frac{7}{3} \\ 0 & 3 & -4 & -7 \end{array} \right]$$

Perfect! We have a leading 1 in the second row.

Step 4: Get a zero below the leading 1 in the second column

We need to eliminate the 3 in the third row, second column. We'll subtract 3 times the second row from the third row:

$$R_3 \rightarrow R_3 - 3R_2: \quad \left[ \begin{array}{rrr|r} 1 & 3 & -2 & 4 \\ 0 & 1 & -\frac{4}{3} & -\frac{7}{3} \\ 0 & 0 & 0 & 0 \end{array} \right]$$

Woah! The entire third row became zeros. This is a significant clue about the nature of our system of equations.

We've successfully transformed the matrix into row-echelon form. Now, let's interpret what this means for the solution of our system.

Interpreting the Row-Echelon Form and Finding the Solution

Okay, so we've diligently applied Gaussian elimination and transformed our augmented matrix into row-echelon form. Here it is again for reference:

$$\left[ \begin{array}{rrr|r} 1 & 3 & -2 & 4 \\ 0 & 1 & -\frac{4}{3} & -\frac{7}{3} \\ 0 & 0 & 0 & 0 \end{array} \right]$$

Now comes the fun part: figuring out what this matrix tells us about the solution to our system of equations. The key thing to notice here is the last row of zeros. This indicates that we have a dependent system, meaning there are infinitely many solutions.

Let's rewrite the matrix back into equation form to make this clearer. The first row corresponds to the equation:

$$x + 3y - 2z = 4$$

The second row corresponds to the equation:

$$y - \frac{4}{3}z = -\frac{7}{3}$$

The third row, being all zeros, doesn't give us any new information.

Since we have three variables (x, y, and z) but only two equations, we'll have one free variable. This means we can choose a value for one variable and express the other variables in terms of it. It's common to choose z as the free variable in this case. Let's set z = t, where t is any real number.

Now we can solve for y in terms of t using the second equation:

$$y - \frac{4}{3}t = -\frac{7}{3}$$

$$y = \frac{4}{3}t - \frac{7}{3}$$

Next, we'll substitute this expression for y and z = t into the first equation to solve for x:

$$x + 3\left(\frac{4}{3}t - \frac{7}{3}\right) - 2t = 4$$

$$x + 4t - 7 - 2t = 4$$

$$x + 2t = 11$$

$$x = -2t + 11$$

So, we've found expressions for x and y in terms of the free variable t. This means the general solution to our system of equations is:

$$x = -2t + 11$$

$$y = \frac{4}{3}t - \frac{7}{3}$$

$$z = t$$

where t can be any real number. This represents an infinite number of solutions, each corresponding to a different value of t. The solutions form a line in three-dimensional space.

In summary, by using Gaussian elimination, we not only found the solution to the system but also determined that it has infinitely many solutions. This is a powerful demonstration of the technique's ability to handle different types of systems, whether they have a unique solution, no solution, or infinitely many solutions.

Conclusion

And there you have it, guys! We've successfully navigated through Gaussian elimination to solve a system of equations. We started by setting up the augmented matrix, then skillfully applied elementary row operations to transform it into row-echelon form. Finally, we interpreted the matrix to find the complete solution, which in this case turned out to be a set of infinitely many solutions parameterized by a free variable.

This exercise highlights the power and versatility of Gaussian elimination. It's not just a method for finding numerical answers; it's a fundamental tool for understanding the nature of linear systems. Whether a system has a unique solution, no solution, or infinitely many solutions, Gaussian elimination can help us uncover the truth.

Remember, the key to mastering Gaussian elimination is practice. Work through different examples, try systems with different numbers of equations and variables, and don't be afraid to make mistakes along the way. Each mistake is a learning opportunity, and with enough practice, you'll become a pro at solving systems of equations using this powerful technique.

So, keep practicing, keep exploring, and keep those linear algebra skills sharp! You've got this!