Solving 4x + 2y + 3z = 42, X + 2y + Z = 20, 2x + 4y + Z = 32 Using Matrix Row Echelon Form
Hey everyone! Today, we're diving into a super cool way to solve systems of linear equations using matrices. Specifically, we're going to tackle a system with three equations and three unknowns, and we'll use the row echelon form method. It might sound intimidating, but trust me, it's a very organized and efficient way to find solutions. Let's break it down step by step.
Understanding the System of Equations
First, let's take a look at the system we're going to solve. We have these three equations:
- 4x + 2y + 3z = 42
- x + 2y + z = 20
- 2x + 4y + z = 32
Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously. One way to do this is by using the method of substitution or elimination, but we're going to level up our game by using matrices. Why matrices? Because they provide a neat and structured way to handle these calculations, especially when you have larger systems of equations.
Before we jump into the matrix operations, it's crucial to grasp the essence of what we're doing. Each equation represents a plane in 3D space, and the solution to the system is the point where all three planes intersect. Think of it like finding the exact spot where three pieces of paper meet if you held them up in the air. This geometrical interpretation can make the algebraic manipulations feel less abstract and more intuitive. This intersection point, if it exists and is unique, gives us the values of x, y, and z that satisfy our system of equations. Sometimes, the planes might not intersect at all (meaning no solution), or they might intersect along a line (meaning infinite solutions). These cases are also fascinating and can be identified using matrix methods, but for now, let's focus on finding a unique solution. The beauty of using matrices is that the process we'll follow – transforming the matrix into row echelon form – is systematic and can reveal whether we have a unique solution, no solution, or infinite solutions. Understanding this underlying principle will not only help you solve this specific system but also equip you with a powerful tool for tackling more complex problems in linear algebra and beyond.
Transforming to an Augmented Matrix
The first thing we need to do is transform our system of equations into an augmented matrix. An augmented matrix is simply a way of representing the system's coefficients and constants in a compact form. We create a matrix where each row represents an equation, and each column represents the coefficients of the variables (x, y, z) and the constants on the right side of the equations. Think of it as extracting the essential numerical information from the equations and arranging it neatly. This step is crucial because it allows us to perform operations on the rows of the matrix, which correspond to manipulating the equations themselves. The goal is to simplify the system without changing the solution.
Here's how we create the augmented matrix from our system:
| 4 2 3 | 42 |
| 1 2 1 | 20 |
| 2 4 1 | 32 |
Notice how the coefficients of x (4, 1, 2) form the first column, the coefficients of y (2, 2, 4) form the second column, the coefficients of z (3, 1, 1) form the third column, and the constants (42, 20, 32) form the last column, separated by a vertical line. This line symbolizes the equals sign in our original equations. The augmented matrix is a powerful tool because it allows us to perform row operations, which are equivalent to manipulating the original equations, in a systematic and organized way. For example, swapping two rows is like swapping two equations, multiplying a row by a constant is like multiplying an equation by a constant, and adding a multiple of one row to another is like adding a multiple of one equation to another. These operations are the key to transforming the matrix into row echelon form, which will reveal the solution to our system. By understanding this initial transformation into an augmented matrix, we've laid the foundation for the next steps in solving the system using row operations.
Achieving Row Echelon Form
Now comes the fun part: transforming our augmented matrix into row echelon form. Row echelon form is a specific arrangement of the matrix that makes it super easy to solve the system of equations. Imagine it as organizing a messy room; once everything is in its place, it's much easier to find what you're looking for. In row echelon form, we want the matrix to have a "stair-step" pattern of leading entries (which are the first non-zero numbers in each row). Specifically, we aim for the following:
- All rows consisting entirely of zeros are at the bottom of the matrix.
- The leading entry (the first non-zero number) in each non-zero row is to the right of the leading entry in the row above it.
- All entries in the column below a leading entry are zeros.
We achieve this form by performing elementary row operations. These operations are the tools we use to rearrange and simplify the matrix. There are three types of elementary row operations:
- Swapping two rows.
- Multiplying a row by a non-zero constant.
- Adding a multiple of one row to another row.
These operations are crucial because they don't change the solution to the original system of equations. Think of them as carefully rearranging the equations without altering their underlying meaning. Each operation corresponds to a valid algebraic manipulation of the original equations. For instance, swapping two rows is like simply reordering the equations, which doesn't affect the solution. Multiplying a row by a constant is the same as multiplying an entire equation by that constant, which also doesn't change the solution. Adding a multiple of one row to another is equivalent to adding a multiple of one equation to another, a fundamental technique in solving systems of equations. The key is to apply these operations strategically to create zeros in the desired locations, gradually transforming the matrix into row echelon form. This process might seem a bit like a puzzle at first, but with practice, you'll develop an intuition for which operations to use and when. By mastering these row operations, you'll be able to tackle a wide range of systems of equations using the matrix method.
Let's apply these operations to our matrix:
| 4 2 3 | 42 |
| 1 2 1 | 20 |
| 2 4 1 | 32 |
Step 1: Swap Row 1 and Row 2 (to get a 1 in the top-left position):
| 1 2 1 | 20 |
| 4 2 3 | 42 |
| 2 4 1 | 32 |
Step 2: Replace Row 2 with Row 2 - 4 * Row 1, and Replace Row 3 with Row 3 - 2 * Row 1 (to get zeros below the leading 1 in the first column):
| 1 2 1 | 20 |
| 0 -6 -1 | -38 |
| 0 0 -1 | -8 |
We now have our matrix in row echelon form! Notice the stair-step pattern of the leading entries (1, -6, and -1). Also, all entries below the leading entries are zeros. This form is incredibly helpful because it allows us to easily solve for the variables using back-substitution.
Back-Substitution to Find the Solution
Once we have our matrix in row echelon form, the final step is to use back-substitution to find the values of x, y, and z. Back-substitution is a technique where we start with the last equation (represented by the last row of the matrix) and solve for one variable. Then, we substitute that value into the equation above it to solve for another variable, and so on, working our way back up the system. Think of it as unwinding a puzzle, starting from the end and working backward to reveal the solution. This method is particularly effective when the matrix is in row echelon form because the structure of the matrix makes it easy to isolate each variable one at a time.
Let's convert our row echelon form matrix back into equations:
- x + 2y + z = 20
- -6y - z = -38
- -z = -8
Now we can use back-substitution:
From equation 3, we get: z = 8
Substitute z = 8 into equation 2:
-6y - 8 = -38 -6y = -30 y = 5
Substitute y = 5 and z = 8 into equation 1:
x + 2(5) + 8 = 20 x + 10 + 8 = 20 x + 18 = 20 x = 2
So, our solution is x = 2, y = 5, and z = 8. We've successfully solved the system of equations using matrix row echelon form and back-substitution!
Verification and Conclusion
It's always a good idea to verify our solution by plugging the values of x, y, and z back into the original equations to make sure they hold true. This step is like double-checking your work on a test; it gives you confidence that your solution is correct. In our case, we found x = 2, y = 5, and z = 8. Let's plug these values into our original equations:
- 4x + 2y + 3z = 42 4(2) + 2(5) + 3(8) = 8 + 10 + 24 = 42 (Correct!)
- x + 2y + z = 20 2 + 2(5) + 8 = 2 + 10 + 8 = 20 (Correct!)
- 2x + 4y + z = 32 2(2) + 4(5) + 8 = 4 + 20 + 8 = 32 (Correct!)
Since our solution satisfies all three equations, we can confidently say that it is the correct solution. We've successfully navigated the world of matrices and row echelon form to solve a system of linear equations. This method is not only efficient but also provides a structured approach to solving more complex systems. Guys, understanding this process is super valuable for various fields, including engineering, computer science, and economics, where solving systems of equations is a common task.
In conclusion, using the matrix row echelon form is a powerful technique for solving systems of linear equations. By transforming the system into an augmented matrix, applying elementary row operations to achieve row echelon form, and using back-substitution, we can systematically find the solution. Remember, practice makes perfect, so try solving different systems of equations using this method to build your skills and confidence. You've got this!
