Solving Linear Equations With Augmented Matrices

by ADMIN 49 views

Hey everyone! Today, we're diving deep into the world of linear equations and augmented matrices. We'll be tackling a specific problem where we're given an augmented matrix and asked to figure out the solution to a system of linear equations. It might sound a bit intimidating at first, but trust me, it's not as scary as it looks. We'll break it down step by step, so even if you're new to this, you'll be able to follow along. So, grab your pencils, and let's get started!

Understanding Augmented Matrices and Linear Equations

Alright, let's start with the basics. What exactly is an augmented matrix, and how does it relate to linear equations? Well, an augmented matrix is simply a way to represent a system of linear equations in a more compact and organized form. Think of it as a shorthand notation that makes it easier to manipulate and solve the equations. Each row in the matrix represents an equation, and each column corresponds to a variable (like x, y, or z). The last column of the matrix contains the constants on the right-hand side of the equations. The vertical line separates the coefficients of the variables from the constants.

So, if we have a system of linear equations, like:

x + 3z = 3 -2x + y - 2z = -15/2 4x + 12z = 12

We can represent it as an augmented matrix like this:

[1033−21−2−152401212]\left[\begin{array}{ccc:c}1 & 0 & 3 & 3 \\-2 & 1 & -2 & -\frac{15}{2} \\4 & 0 & 12 & 12\end{array}\right]

See how each row in the matrix corresponds to an equation? The first row represents the equation x + 3z = 3, the second row represents -2x + y - 2z = -15/2, and the third row represents 4x + 12z = 12. The augmented matrix provides a clear and organized way to see the system of equations. Our primary goal is to solve the system of linear equations by using the augmented matrix representation, we aim to find the values of the variables x, y, and z that satisfy all the equations simultaneously.

Solving the System of Linear Equations

Now, let's get to the fun part: solving the system of linear equations using the augmented matrix. The primary goal is to transform the augmented matrix into a simpler form, like row-echelon form or reduced row-echelon form. This can be achieved through elementary row operations, which are operations that do not change the solution set of the system. The common elementary row operations are swapping two rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another row. By applying these operations strategically, we can systematically eliminate variables and isolate them. For instance, we may start by eliminating x from the second and third equations.

Let's take a look at the given augmented matrix again:

[1033−21−2−152401212]\left[\begin{array}{ccc:c}1 & 0 & 3 & 3 \\-2 & 1 & -2 & -\frac{15}{2} \\4 & 0 & 12 & 12\end{array}\right]

To eliminate x from the second row, we add 2 times the first row to the second row. This gives us:

[1033014−32401212]\left[\begin{array}{ccc:c}1 & 0 & 3 & 3 \\0 & 1 & 4 & -\frac{3}{2} \\4 & 0 & 12 & 12\end{array}\right]

Next, to eliminate x from the third row, we subtract 4 times the first row from the third row. This yields:

[1033014−320000]\left[\begin{array}{ccc:c}1 & 0 & 3 & 3 \\0 & 1 & 4 & -\frac{3}{2} \\0 & 0 & 0 & 0\end{array}\right]

Now we've got a much simpler matrix. Let's analyze this simplified matrix to extract the solutions. From the first row, we have x + 3z = 3, which can be rearranged to x = 3 - 3z. The second row provides us with y + 4z = -3/2, which rearranges to y = -3/2 - 4z. The third row, where all the entries are zeros, means we have 0 = 0, which is always true and provides no information about the variables.

Since we have two equations and three unknowns, the system has infinitely many solutions. This is because we can choose a value for z and then calculate the corresponding values for x and y. Therefore, we can express the solution in terms of z, letting z = t, where t is any real number. Then the solution set is x = 3 - 3t, y = -3/2 - 4t, z = t.

Step-by-Step Breakdown

Let's walk through the steps again, this time with a bit more detail:

  1. Understand the Augmented Matrix: First, make sure you understand how the augmented matrix represents the system of linear equations. Each row is an equation, and each column corresponds to a variable. The last column is the constant terms.
  2. Apply Elementary Row Operations: Use elementary row operations (swapping rows, multiplying a row by a constant, and adding a multiple of one row to another) to transform the matrix. The aim is to get the matrix into row-echelon form or reduced row-echelon form.
  3. Solve for the Variables: Once the matrix is in a simpler form, you can easily solve for the variables. If you have a unique solution, you'll get values for each variable. If you have infinitely many solutions, you'll express the variables in terms of a parameter (like t in our example).
  4. Check Your Answer: Always check your solution by substituting the values of the variables back into the original equations to ensure they satisfy all the equations.

Tips and Tricks

Here are some helpful tips to keep in mind:

  • Be Organized: Keep your work neat and organized. This will help you avoid making mistakes, especially when dealing with multiple row operations.
  • Double-Check Your Arithmetic: Carefully check your calculations at each step. Even a small error can lead to a wrong answer.
  • Practice, Practice, Practice: The more you practice, the more comfortable you'll become with solving systems of linear equations using augmented matrices.
  • Use Technology: If allowed, use calculators or software to perform matrix operations. This can save you time and help you check your work.
  • Understand the Forms: Familiarize yourself with row-echelon form and reduced row-echelon form. Knowing these forms will guide you in simplifying the matrix.

Conclusion

And there you have it! We've successfully solved a system of linear equations using an augmented matrix. Remember, this is a fundamental concept in linear algebra, and it has many applications in various fields, including engineering, economics, and computer science. By understanding the basics and practicing regularly, you'll be able to tackle these problems with confidence. Keep up the great work, and happy solving!

I hope you enjoyed this guide. Let me know if you have any questions in the comments below. Happy learning, and I'll catch you in the next one! Bye for now!