Gauss-Jordan Elimination: Solve System Of Equations

Hey guys! Let's dive into solving systems of equations using the Gauss-Jordan elimination method. It's a super powerful technique that helps us find solutions, even when things get a little complex. We'll break down the process step-by-step, so don't worry if it sounds intimidating at first. We're going to tackle a specific system of equations today, and by the end, you'll be feeling confident in your ability to solve these problems. Let's get started!

Understanding Gauss-Jordan Elimination

Gauss-Jordan elimination is essentially a systematic way to transform a system of linear equations into a simpler, equivalent system that's much easier to solve. At its heart, this method relies on performing elementary row operations on an augmented matrix. But what exactly does that mean? Let's break it down further. The core idea revolves around manipulating the coefficients and constants of our equations in a structured manner until we reach a form where the solution becomes crystal clear.

First off, we represent our system of equations as an augmented matrix. This matrix is like a condensed version of our equations, where each row represents an equation, and each column corresponds to the coefficients of a particular variable (plus an additional column for the constants on the right side of the equations). Think of it as organizing all the key numbers from our equations into a neat table. This augmented matrix then becomes our playground for the Gauss-Jordan elimination process. We'll be performing operations on it to achieve a specific form, a form that directly reveals the solution to our system.

Now, what are these "elementary row operations" we keep mentioning? These are the secret moves in our Gauss-Jordan toolkit. There are three main types:

1. Swapping two rows: Imagine you have two equations in your system, and you decide to switch their order. That's essentially what this row operation does. It doesn't change the solution to the system, just the order in which the equations are presented.
2. Multiplying a row by a non-zero constant: This is like scaling an entire equation. You can multiply all the terms in an equation by the same number without affecting the overall solution. For instance, you might multiply a row by 2 to get rid of a fraction or simplify the coefficients.
3. Adding a multiple of one row to another: This is where things get interesting. It allows us to eliminate variables systematically. By adding a suitable multiple of one row to another, we can create zeros in specific positions in the matrix, gradually working towards our desired form.

The goal of Gauss-Jordan elimination is to use these row operations to transform the augmented matrix into what's called reduced row-echelon form. This form has a very distinctive look. The left side of the matrix (the part corresponding to the coefficients) should resemble an identity matrix – a square matrix with 1s on the main diagonal and 0s everywhere else. The right side of the matrix (the part corresponding to the constants) then directly gives us the solution to the system of equations. It's like the grand finale of our mathematical dance, where the solution is revealed in a clear and elegant way.

So, in essence, Gauss-Jordan elimination is a powerful and systematic technique for solving systems of linear equations. It might seem like a lot of steps at first, but with practice, it becomes a familiar and reliable method for tackling these types of problems. Remember, the key is to focus on those elementary row operations and use them strategically to transform the matrix into reduced row-echelon form. Once you get the hang of it, you'll be solving systems of equations like a pro! We’ll see how this unfolds with our example in just a bit. Stay tuned!

Setting up the Augmented Matrix

Okay, guys, let's get practical! The first step in using Gauss-Jordan elimination is to translate our system of equations into an augmented matrix. This matrix is basically a compact way of representing our equations, making it easier to manipulate them. Remember those coefficients and constants we talked about? They're about to take center stage!

Our system of equations is:

x - y + 6z = 5
x - y + 7z = 7

To form the augmented matrix, we'll create a rectangular array. Each row will represent one of our equations, and each column will correspond to the coefficients of a specific variable, plus the constants on the right-hand side. It's like creating a table where we neatly organize all the numerical information from our equations. Think of it as a mathematical spreadsheet, ready for some action!

Let's break it down column by column:

• The first column will hold the coefficients of the x variable.
• The second column will hold the coefficients of the y variable.
• The third column will hold the coefficients of the z variable.
• And finally, the last column, separated by a vertical line, will hold the constants on the right side of the equations.

So, looking at our first equation, x - y + 6z = 5, we have:

• Coefficient of x: 1
• Coefficient of y: -1
• Coefficient of z: 6
• Constant: 5

Similarly, for the second equation, x - y + 7z = 7, we have:

• Coefficient of x: 1
• Coefficient of y: -1
• Coefficient of z: 7
• Constant: 7

Now, we'll arrange these numbers into our augmented matrix:

[ 1  -1   6 | 5 ]
[ 1  -1   7 | 7 ]

See how each row corresponds to an equation, and the vertical line separates the coefficients from the constants? This is our augmented matrix, ready for the next stage of the Gauss-Jordan elimination process. It might seem like a simple step, but setting up the matrix correctly is crucial for getting the right answer. It's the foundation upon which we'll build our solution. Now that we have our matrix, we're ready to start performing row operations and transforming it into that magical reduced row-echelon form we talked about earlier. It's like preparing our ingredients before we start cooking – we've got everything in place, and now the fun begins! Next up, we'll start manipulating the rows to get closer to our solution. Keep going, we’re making progress!

Performing Row Operations

Alright, team, we've got our augmented matrix set up, and now it's time to roll up our sleeves and get into the real work: performing row operations. This is where we strategically manipulate the rows of the matrix to transform it into reduced row-echelon form. Remember those three elementary row operations we discussed? Now we're going to put them into action! The goal here is to create a matrix that has a clear pattern of 1s and 0s, making the solution to our system of equations jump right out at us. Think of it like solving a puzzle, where each row operation is a move that gets us closer to the final picture.

Our augmented matrix is:

[ 1  -1   6 | 5 ]
[ 1  -1   7 | 7 ]

Our first goal is to get a 0 in the bottom left corner of the matrix (the position where we currently have a 1). To do this, we can subtract the first row from the second row. This is an example of using the row operation: "Adding a multiple of one row to another." In this case, we're adding -1 times the first row to the second row. The notation for this operation is often written as R2 -> R2 - R1, which means "replace row 2 with row 2 minus row 1". It’s like giving the matrix a little makeover!

Let's perform the operation:

R2 -> R2 - R1

This means we'll subtract each element in the first row from the corresponding element in the second row:

• New element in R2, column 1: 1 - 1 = 0
• New element in R2, column 2: -1 - (-1) = 0
• New element in R2, column 3: 7 - 6 = 1
• New element in R2, constant column: 7 - 5 = 2

Our updated matrix now looks like this:

[ 1  -1   6 | 5 ]
[ 0   0   1 | 2 ]

Awesome! We've got a 0 in the bottom left corner. Now, we want to get a 0 in the first row, third column (where we currently have a 6). To do this, we can subtract 6 times the second row from the first row. This operation can be written as R1 -> R1 - 6R2. See how we're using those row operations strategically to create the zeros we need? It’s all about planning our moves!

Let's perform this operation:

R1 -> R1 - 6R2

This means we'll subtract 6 times each element in the second row from the corresponding element in the first row:

• New element in R1, column 1: 1 - 6(0) = 1
• New element in R1, column 2: -1 - 6(0) = -1
• New element in R1, column 3: 6 - 6(1) = 0
• New element in R1, constant column: 5 - 6(2) = -7

Our matrix is now:

[ 1  -1   0 | -7 ]
[ 0   0   1 | 2 ]

Look at that! We're getting closer and closer to reduced row-echelon form. We have a diagonal of 1s with a 0 below the first 1 and a 0 above the second 1 in the third column. At this point, we've done all we can with row operations. Notice that we don't have a leading 1 in the second row, second column. This tells us something important about the system of equations – it has infinitely many solutions! We'll interpret this result in the next section. The key takeaway here is that performing row operations systematically allows us to simplify the matrix and reveal the nature of the solutions. We're like mathematical detectives, uncovering the hidden truths within the equations. Keep it up, you're doing great!

Interpreting the Solution

Okay, everyone, we've successfully transformed our augmented matrix using Gauss-Jordan elimination! We've manipulated the rows to get it into a simplified form, and now comes the exciting part: interpreting what the matrix is telling us. This is where we translate the numbers back into the language of equations and figure out the solution (or solutions) to our original system. It's like deciphering a code, where the matrix holds the key to unlocking the answer.

Our final matrix looks like this:

[ 1  -1   0 | -7 ]
[ 0   0   1 | 2 ]

Remember, each row represents an equation. Let's rewrite these rows as equations:

• Row 1: 1x - 1y + 0z = -7 which simplifies to x - y = -7
• Row 2: 0x + 0y + 1z = 2 which simplifies to z = 2

So, we've directly found that z = 2. But what about x and y? Notice that the first equation, x - y = -7, has two variables. This means we can't solve for a unique value for both x and y. This is a sign that our system has infinitely many solutions, which occurs when we have fewer equations than variables (after the matrix is in reduced row-echelon form) or when we encounter rows of all zeros. It’s like having a puzzle with missing pieces – we can’t get a single, definitive answer.

To express these infinite solutions, we need to introduce a parameter. A parameter is just a variable that we use to describe the range of possible solutions. In this case, let's let y be our parameter. We'll call it t, so y = t. This is like saying, "Okay, y can be any number, so let's just call it t for now."

Now, we can substitute y = t into our equation x - y = -7:

x - t = -7

Solving for x, we get:

x = t - 7

So, we've expressed x in terms of our parameter t. We already know that z = 2. Therefore, our solution can be written as:

• x = t - 7
• y = t
• z = 2

This is a parametric solution. It tells us that for any value we choose for t, we can find corresponding values for x and y that satisfy the original system of equations. It's like a recipe for solutions – we plug in a value for t, and it spits out a solution for us. For example:

• If t = 0, then x = -7, y = 0, and z = 2
• If t = 1, then x = -6, y = 1, and z = 2
• If t = -1, then x = -8, y = -1, and z = 2

And so on! There are infinitely many solutions, each corresponding to a different value of t. So, to summarize, our Gauss-Jordan elimination led us to a matrix that revealed a dependent system with infinitely many solutions. We then used a parameter to express these solutions in a general form. This process of interpreting the matrix is crucial for understanding the nature of the solutions to our original system of equations. We've not only solved the problem but also gained insight into the relationship between the equations. You've tackled a challenging problem and come out on top! Great work, guys!

Final Solution

Okay, let's bring it all together and present our final solution in a clear and concise way. We've gone through the entire Gauss-Jordan elimination process, from setting up the augmented matrix to performing row operations and interpreting the results. Now, it's time to state the answer in a way that's easy to understand. Think of this as the grand finale, where we showcase our hard work and present the fruit of our labor. We started with a system of equations, navigated through the steps of Gauss-Jordan elimination, and now we're ready to declare victory!

Remember our original system of equations?

x - y + 6z = 5
x - y + 7z = 7

And after performing Gauss-Jordan elimination, we arrived at the following parametric solution:

• x = t - 7
• y = t
• z = 2

Where t is a parameter that can take on any real value. This is how we express the infinitely many solutions to our system. We can't list out every possible solution, but we can provide a formula that generates them. It’s like giving someone the key to a treasure chest – they can unlock as many solutions as they want!

So, our final answer is:

(x, y, z) = (t - 7, t, 2)

This notation represents the set of all solutions to the system. For any value of t, we get a valid solution. This is a concise and elegant way to express the infinite solutions we found. This final step is important because it clearly communicates our findings. We're not just showing the steps we took, but also the ultimate result of our efforts. It's like presenting a finished masterpiece after hours of work – we want to make sure it's displayed in the best possible light.

To recap, we used Gauss-Jordan elimination to transform the system of equations into a simpler form. We identified that the system was dependent, meaning it had infinitely many solutions. We then introduced a parameter, t, to express these solutions in a general way. Finally, we presented the solution in a clear and standard notation. You've successfully navigated the world of Gauss-Jordan elimination and solved a system with infinitely many solutions. Give yourselves a pat on the back – you've earned it! Remember, practice makes perfect, so keep tackling these problems, and you'll become a master of linear algebra in no time. You got this!