Solve Linear Equations By Matrix Method: Step-by-Step

Hey guys! Today, we're diving into the fascinating world of linear equations and how to solve them using the matrix method. This approach is super powerful, especially when you're dealing with systems of equations that have multiple variables. We'll walk through a specific example step-by-step, making sure you understand each part of the process. So, let's get started and make solving these equations a breeze!

The Challenge: A System of Linear Equations

Let's start with the system of equations we need to solve:

x + 2y - z = 4

x - y + 3z = -1

-2y = 5

Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously. The matrix method provides a structured way to tackle this. It involves representing the system of equations in matrix form and then using matrix operations to find the solution. The beauty of this method is its systematic approach, which helps prevent errors and makes the solution process more organized. First, we'll convert this system into a matrix equation. This involves creating three matrices: the coefficient matrix (A), the variable matrix (X), and the constant matrix (B). The coefficient matrix consists of the coefficients of the variables in the equations. The variable matrix is a column matrix containing the variables themselves, and the constant matrix contains the constants on the right-hand side of the equations. This initial setup is crucial because it transforms the algebraic problem into a matrix problem, which can then be solved using established matrix techniques. Once the matrix equation is set up, we can use methods like Gaussian elimination or matrix inversion to solve for the variable matrix, which will give us the values of x, y, and z. The matrix method not only provides a solution but also gives insights into the nature of the system, such as whether the system has a unique solution, infinitely many solutions, or no solution at all. This makes it a versatile tool in various fields, including engineering, economics, and computer science.

Step 1: Representing the System as a Matrix Equation

First, we need to represent our system of equations in matrix form. This involves creating three matrices:

• Coefficient Matrix (A): This matrix contains the coefficients of the variables (x, y, and z) from our equations.
• Variable Matrix (X): This matrix contains the variables we're trying to solve for.
• Constant Matrix (B): This matrix contains the constants on the right side of the equations.

For our system, these matrices look like this:

A = | 1 2 -1 |

| 1 -1  3 |

| 0 -2  0 |

X = | x |

| y |

| z |

B = | 4 |

| -1 |

| 5 |

Now, we can write our system of equations in matrix form as:

AX = B

This matrix equation represents our original system in a compact and organized way. This representation is fundamental to solving the system using matrix methods. The coefficient matrix (A) encapsulates the relationships between the variables, while the variable matrix (X) holds the unknowns we aim to find. The constant matrix (B) represents the results of the linear combinations of the variables. By expressing the system in this form, we can apply various matrix operations to isolate the variable matrix and determine the values of x, y, and z. For instance, we can use techniques like Gaussian elimination, LU decomposition, or matrix inversion to solve for X. Each of these methods involves transforming the coefficient matrix into a simpler form (such as an identity matrix or an upper triangular matrix) while performing corresponding operations on the constant matrix. The resulting transformed matrices then allow us to easily read off the values of the variables. Moreover, the matrix representation provides a clear framework for analyzing the properties of the system, such as its consistency and the uniqueness of solutions. This makes the matrix method a powerful tool not only for solving linear equations but also for understanding the underlying mathematical structure of the system.

Step 2: Solving for the Variables

To solve for the variables (x, y, and z), we can use several methods, such as Gaussian elimination, matrix inversion, or Cramer's rule. For this example, let's use Gaussian elimination, which involves transforming the augmented matrix [A|B] into row-echelon form or reduced row-echelon form. The augmented matrix is formed by appending the constant matrix (B) to the coefficient matrix (A). This combination allows us to perform row operations on both matrices simultaneously, streamlining the solution process. Gaussian elimination systematically eliminates variables from the equations by using elementary row operations: swapping rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another. These operations are applied in a way that gradually transforms the coefficient matrix into an upper triangular form, where all entries below the main diagonal are zero. This simplifies the system of equations, making it easier to solve for the variables using back-substitution. If we go further and transform the matrix into reduced row-echelon form, we obtain an identity matrix on the left side, directly revealing the solution in the rightmost column. This method is particularly efficient and provides a clear pathway to the solution, making it a fundamental technique in linear algebra. Gaussian elimination not only solves the system but also provides valuable information about the system's nature, such as whether it has a unique solution, infinitely many solutions, or no solution at all.

Our augmented matrix [A|B] looks like this:

| 1 2 -1 | 4 |

| 1 -1 3 | -1 |

| 0 -2 0 | 5 |

Now, let's perform row operations to get it into row-echelon form:

1. Subtract row 1 from row 2 (R2 = R2 - R1):

| 1 2 -1 | 4 |

| 0 -3 4 | -5 |

| 0 -2 0 | 5 |

2. Multiply row 3 by -1/2 (R3 = R3 * -1/2):

| 1 2 -1 | 4 |

| 0 -3 4 | -5 |

| 0 1 0 | -5/2 |

3. Swap row 2 and row 3 (R2 <-> R3):

| 1 2 -1 | 4 |

| 0 1 0 | -5/2 |

| 0 -3 4 | -5 |

4. Add 3 times row 2 to row 3 (R3 = R3 + 3R2):

| 1 2 -1 | 4 |

| 0 1 0 | -5/2 |

| 0 0 4 | -25/2 |

5. Divide row 3 by 4 (R3 = R3 / 4):

| 1 2 -1 | 4 |

| 0 1 0 | -5/2 |

| 0 0 1 | -25/8 |

Now our matrix is in row-echelon form. We can now use back-substitution to solve for the variables. Back-substitution is a straightforward method that leverages the triangular structure of the matrix to solve for the variables one by one. Starting from the last row, which gives us the value of the last variable directly, we substitute this value into the equation represented by the second-to-last row to solve for the second-to-last variable. This process continues up the rows, with each variable being solved using the values of the variables already found. Back-substitution is highly efficient and minimizes the chance of errors, as it avoids complex algebraic manipulations. It is a staple technique in solving systems of linear equations, particularly after Gaussian elimination has transformed the system into a simpler, triangular form. The ease and reliability of back-substitution make it an essential tool for anyone working with linear systems.

Step 3: Back-Substitution

From the row-echelon form, we can easily find the values of z, y, and x:

• From the third row: z = -25/8
• From the second row: y = -5/2
• From the first row: x + 2(-5/2) - (-25/8) = 4 => x - 5 + 25/8 = 4 => x = 4 + 5 - 25/8 => x = 47/8

So, the solution is:

x = 47/8

y = -5/2

z = -25/8

Step 4: Verifying the Solution

To make sure we've got the correct solution, it's always a good idea to plug our values back into the original equations and see if they hold true. This step is crucial for catching any errors that might have occurred during the solution process. Substituting the values back into the original equations provides a direct check of the solution's validity. If the equations hold true with the substituted values, it confirms that the solution is correct. This verification step ensures that the algebraic manipulations and row operations were performed accurately. In addition to confirming the correctness of the solution, this step also reinforces understanding of the relationship between the variables and the equations. It provides a tangible connection between the abstract algebraic solution and the concrete equations it is meant to satisfy. Furthermore, this practice is especially important when dealing with complex systems or when using computational tools, as it can help identify input errors or algorithmic issues. By verifying the solution, we gain confidence in the result and ensure that it accurately represents the solution to the original problem.

Let's check our solution:

1. x + 2y - z = 47/8 + 2(-5/2) - (-25/8) = 47/8 - 5 + 25/8 = (47 - 40 + 25)/8 = 32/8 = 4 (Correct!)
2. x - y + 3z = 47/8 - (-5/2) + 3(-25/8) = 47/8 + 5/2 - 75/8 = (47 + 20 - 75)/8 = -8/8 = -1 (Correct!)
3. -2y = -2(-5/2) = 5 (Correct!)

Our solution checks out! We've successfully solved the system of equations using the matrix method.

Conclusion

And there you have it! We've tackled a system of linear equations using the matrix method, walking through each step from setting up the matrices to verifying our solution. This method is a powerful tool in your mathematical arsenal, especially when dealing with more complex systems. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a pro in no time. Keep up the great work, and happy solving!