Solving 3x3 System Of Equations Step-by-Step Guide
This comprehensive guide will walk you through the process of solving a system of three linear equations with three unknowns. We will use a step-by-step approach that combines the elimination method and substitution method to find the values of x, y, and z that satisfy all three equations simultaneously. Solving systems of linear equations is a fundamental concept in mathematics and has wide-ranging applications in various fields, including engineering, physics, economics, and computer science. When faced with a system of equations, the goal is to find the unique set of values for the variables that make all the equations true. This process often involves manipulating the equations algebraically to isolate variables or eliminate them, thereby simplifying the system until a solution can be readily obtained. By mastering these techniques, you'll be well-equipped to tackle more complex mathematical problems and real-world scenarios that rely on the solution of linear systems. This article aims to provide a clear and structured approach to solving these systems, ensuring you grasp the underlying principles and methods effectively.
Problem
Consider the following system of three linear equations:
x + 2y - z = -3
2x - y + z = 5
x - y + z = 4
Our goal is to find the values of x, y, and z that satisfy all three equations. Solving this system involves a systematic approach to eliminate variables and isolate the unknowns. The process begins by choosing a pair of equations to eliminate one variable. This can be achieved by multiplying one or both equations by suitable constants so that the coefficients of the variable to be eliminated are equal in magnitude but opposite in sign. Once the equations are modified, they can be added together, resulting in a new equation with one less variable. This step is repeated with different pairs of equations until the system is reduced to two equations with two variables, which can then be solved using similar elimination or substitution techniques. By carefully applying these methods, we can determine the unique values of x, y, and z that solve the system.
Method 1: Elimination Method
The elimination method involves strategically adding or subtracting multiples of equations to eliminate variables. This method is particularly effective when dealing with systems of equations where coefficients of one variable are easily made equal or opposite. The core idea is to manipulate the equations so that when they are added together, one of the variables cancels out, leaving a simpler equation with fewer unknowns. This process can be repeated to eliminate additional variables until the system is reduced to a single equation with a single variable, which can be easily solved. Once the value of one variable is found, it can be substituted back into the other equations to find the values of the remaining variables. The elimination method is a powerful tool for solving linear systems and provides a systematic way to find the solution by reducing the complexity of the problem step by step. This approach is not only efficient but also helps in understanding the structure and relationships within the system of equations.
Step 1: Eliminate z from the first two equations.
Add the first and second equations:
(x + 2y - z) + (2x - y + z) = -3 + 5
3x + y = 2 ----(4)
By adding the first and second equations, we have successfully eliminated the variable z, resulting in a new equation that involves only x and y. This step is a critical part of the elimination method, which aims to simplify the system of equations by reducing the number of variables in each equation. The resulting equation, 3x + y = 2, is a linear equation in two variables and provides a crucial link towards solving for x and y. This equation, along with another equation involving x and y, will form a smaller system that can be solved using similar techniques, such as elimination or substitution. The elimination of z in this step demonstrates the power of the elimination method in systematically reducing the complexity of the problem, making it easier to find the solution for the original system of equations.
Step 2: Eliminate z from the first and third equations.
Add the first and third equations:
(x + 2y - z) + (x - y + z) = -3 + 4
2x + y = 1 ----(5)
Similar to the previous step, adding the first and third equations eliminates the variable z, leading to another equation involving only x and y. This step reinforces the strategy of the elimination method, where variables are systematically removed to simplify the system. The new equation, 2x + y = 1, now forms a pair with the equation obtained in the previous step (3x + y = 2), creating a 2x2 system of equations. This smaller system is significantly easier to solve and can be tackled using various methods, such as substitution or further elimination. The process of reducing a 3x3 system to a 2x2 system is a key milestone in solving linear systems, as it allows us to focus on fewer variables and equations, making the solution more accessible. The successful elimination of z in both steps demonstrates the effectiveness of this approach in streamlining the problem-solving process.
Step 3: Solve the system of two equations ((4) and (5)) for x and y.
Subtract equation (5) from equation (4):
(3x + y) - (2x + y) = 2 - 1
x = 1
By subtracting equation (5) from equation (4), we have successfully eliminated the variable y and directly solved for x. This step highlights the elegance of the elimination method, where strategic operations can isolate a single variable, leading to its value. The result, x = 1, is a significant milestone in solving the system of equations, as it provides the first concrete value for one of the unknowns. With the value of x now known, we can substitute it back into one of the equations involving x and y to find the value of y. This process of substitution is a common technique in solving systems of equations and allows us to leverage the value of one variable to find the others. The ability to isolate and solve for a variable in this manner underscores the power and efficiency of the elimination method in tackling linear systems.
Step 4: Substitute x = 1 into equation (5) to find y.
2(1) + y = 1
2 + y = 1
y = -1
Having found the value of x, substituting it back into equation (5) allows us to solve for y. This step demonstrates the effectiveness of the substitution method, where known values are used to simplify equations and find the remaining unknowns. By plugging x = 1 into 2x + y = 1, we obtain a simple equation in terms of y, which can be easily solved to find y = -1. This process of substitution is a crucial part of solving systems of equations, as it allows us to leverage the values we have already found to determine the values of the remaining variables. The successful determination of y = -1 is another significant step towards solving the entire system, as we now have the values for two of the three variables. With x and y known, we can proceed to find the value of z by substituting these values into one of the original equations.
Step 5: Substitute x = 1 and y = -1 into the first original equation to find z.
1 + 2(-1) - z = -3
1 - 2 - z = -3
-1 - z = -3
z = 2
With the values of x and y determined, the final step involves substituting these values back into one of the original equations to solve for z. This step showcases the culmination of the elimination and substitution methods, where previously found values are used to unravel the remaining unknowns. By substituting x = 1 and y = -1 into the first original equation, x + 2y - z = -3, we obtain a simple equation in terms of z. Solving this equation yields z = 2, which completes the solution for the system of equations. This final step underscores the importance of systematically applying the elimination and substitution methods to arrive at a complete solution. With the values of x, y, and z now known, we have successfully solved the system of equations.
Solution
Therefore, the solution to the system of equations is:
x = 1
y = -1
z = 2
We have successfully found the values of x, y, and z that satisfy all three equations simultaneously. The solution x = 1, y = -1, and z = 2 represents the unique point in three-dimensional space where the planes defined by the three equations intersect. This solution is the result of a systematic application of the elimination and substitution methods, which allowed us to reduce the complexity of the system and isolate each variable. The ability to solve systems of linear equations is a fundamental skill in mathematics and has wide-ranging applications in various fields, including engineering, physics, economics, and computer science. This comprehensive step-by-step solution provides a clear understanding of the process and highlights the importance of methodical problem-solving techniques.
Method 2: Matrix Method (Optional)
The matrix method provides a powerful and efficient approach to solving systems of linear equations, especially when dealing with larger systems. This method leverages the principles of linear algebra to represent the system in a compact matrix form, allowing for systematic manipulation and solution. The core idea is to represent the coefficients of the variables and the constants as matrices and then use matrix operations, such as Gaussian elimination or matrix inversion, to find the solution. The matrix method is particularly advantageous because it can be easily implemented using computer software and programming languages, making it a practical tool for solving complex systems of equations in various applications, including engineering, physics, and computer science. Understanding the matrix method not only enhances problem-solving skills but also provides a deeper insight into the mathematical structure of linear systems and their solutions. This approach is widely used in both theoretical and applied mathematics, making it an essential tool for anyone working with linear systems.
Step 1: Represent the system as an augmented matrix.
The system of equations can be represented as the following augmented matrix:
[ 1 2 -1 | -3 ]
[ 2 -1 1 | 5 ]
[ 1 -1 1 | 4 ]
The augmented matrix is a compact representation of the system of equations, where the coefficients of the variables and the constants are arranged in a rectangular array. This matrix form allows us to apply systematic matrix operations to solve the system efficiently. The left side of the matrix contains the coefficients of the variables x, y, and z, while the rightmost column represents the constants on the right-hand side of the equations. This representation is a crucial step in the matrix method, as it transforms the system of equations into a format that can be easily manipulated using matrix algebra. The augmented matrix provides a clear and organized way to visualize the system and apply operations such as row reduction to find the solution. This approach is particularly useful for larger systems of equations, where the matrix representation simplifies the problem-solving process.
Step 2: Perform row operations to get the matrix in row-echelon form.
Apply the following row operations:
- R2 -> R2 - 2R1
- R3 -> R3 - R1
The matrix becomes:
[ 1 2 -1 | -3 ]
[ 0 -5 3 | 11 ]
[ 0 -3 2 | 7 ]
Row operations are fundamental to the matrix method for solving systems of linear equations. These operations involve manipulating the rows of the augmented matrix in a systematic way to simplify the matrix and ultimately solve for the variables. The goal is to transform the matrix into row-echelon form, where the leading coefficient (the first non-zero entry) of each row is to the right of the leading coefficient of the row above it, and all entries in a column below a leading coefficient are zero. The row operations include swapping two rows, multiplying a row by a non-zero constant, and adding a multiple of one row to another. These operations do not change the solution of the system and allow us to systematically eliminate variables. By applying these operations, we can reduce the matrix to a form where the solution can be easily obtained through back-substitution. The row-echelon form is a crucial intermediate step in the matrix method, making the solution process more manageable and efficient.
Step 3: Continue row operations.
Apply the following row operation:
- R3 -> 5R3 - 3R2
The matrix becomes:
[ 1 2 -1 | -3 ]
[ 0 -5 3 | 11 ]
[ 0 0 1 | 2 ]
Continuing the row operations, we further transform the matrix towards its row-echelon form. This step often involves more complex manipulations, such as multiplying rows by constants and adding multiples of one row to another, to create zeros in strategic locations within the matrix. The goal is to simplify the system as much as possible, making it easier to solve for the variables. In this particular step, the row operation 5R3 - 3R2 is applied, which eliminates the entry in the second column of the third row, bringing the matrix closer to its row-echelon form. This systematic elimination of entries is a key aspect of the matrix method and allows us to isolate the variables one by one. The resulting matrix has a triangular structure, which makes it straightforward to solve for the variables using back-substitution. The progression through these row operations demonstrates the power of matrix manipulation in solving linear systems.
Step 4: Convert back to equations and solve using back-substitution.
The matrix now represents the following system of equations:
x + 2y - z = -3
-5y + 3z = 11
z = 2
Converting the row-echelon form of the matrix back into a system of equations allows us to use back-substitution to find the values of the variables. This step bridges the gap between the matrix representation and the original system of equations, making the solution process more intuitive. The row-echelon form has a triangular structure, where the last equation directly gives the value of one variable (z in this case). With the value of z known, we can substitute it into the second equation to solve for y. This process is repeated, substituting the known values into the preceding equations until all variables are solved. Back-substitution is an efficient way to solve systems of equations that are in row-echelon form, as it systematically works backwards from the simplest equation to the most complex. This method highlights the advantage of using row operations to transform the matrix into a form that facilitates easy solution.
Step 5: Solve for y and x.
Substitute z = 2 into -5y + 3z = 11:
-5y + 3(2) = 11
-5y + 6 = 11
-5y = 5
y = -1
Substitute y = -1 and z = 2 into x + 2y - z = -3:
x + 2(-1) - 2 = -3
x - 2 - 2 = -3
x - 4 = -3
x = 1
By using back-substitution, we systematically solve for the remaining variables, y and x. This step demonstrates the power of leveraging known values to simplify equations and find the unknowns. Starting with the equation -5y + 3z = 11, we substitute z = 2 to find y = -1. This value is then used in conjunction with z = 2 in the equation x + 2y - z = -3 to solve for x, resulting in x = 1. This sequential substitution is a hallmark of the back-substitution method and allows us to efficiently unravel the solution. The process highlights the interconnectedness of the variables in the system of equations and how finding one variable can lead to the determination of others. This final step completes the solution process, providing the values for all variables in the system.
Solution
Therefore, the solution to the system of equations is:
x = 1
y = -1
z = 2
We have successfully solved the system of equations using the matrix method, arriving at the same solution as with the elimination method. This consistency underscores the reliability of both methods in solving linear systems. The matrix method provides a systematic and efficient way to manipulate the equations, especially for larger systems, while the back-substitution process allows us to easily find the values of the variables once the matrix is in row-echelon form. The solution x = 1, y = -1, and z = 2 satisfies all three original equations, confirming its validity. This comprehensive solution process demonstrates the power of matrix algebra in solving linear systems and highlights its importance in various scientific and engineering applications.
Conclusion
In this article, we explored two methods for solving a system of three linear equations with three variables: the elimination method and the matrix method. Both methods provide a systematic approach to finding the values of x, y, and z that satisfy all three equations simultaneously. The elimination method involves strategically adding or subtracting multiples of equations to eliminate variables, while the matrix method uses row operations on an augmented matrix to achieve row-echelon form, followed by back-substitution. Understanding and mastering these methods is crucial for solving various mathematical problems and real-world applications. The ability to solve systems of linear equations is a fundamental skill in mathematics and has wide-ranging applications in various fields, including engineering, physics, economics, and computer science. By understanding these methods, you can effectively tackle similar problems and gain a deeper appreciation for linear algebra.
The solution to the system
x + 2y - z = -3
2x - y + z = 5
x - y + z = 4
is x = 1, y = -1, and z = 2. This solution can be verified by substituting these values back into the original equations and confirming that they hold true. The systematic approach used in both methods ensures that the solution is accurate and reliable. Whether you choose to use the elimination method or the matrix method, the key is to apply the steps consistently and carefully to arrive at the correct solution. Solving systems of linear equations is a powerful tool in mathematics, and with practice, you can become proficient in using these methods to solve a wide range of problems.
