Solving AX=b And Geometric Interpretation Of The Solution
Introduction
In this article, we delve into the process of solving a system of linear equations represented in the matrix form AX = b. Specifically, we will focus on the given system where the matrix A is a 3x3 matrix, X is a column vector of variables, and b is a constant column vector. Our primary goal is to find the solution vector X that satisfies the equation. Furthermore, we will provide a geometric interpretation of the solution, visualizing the equations as planes in three-dimensional space and understanding how their intersection leads to the solution.
The system of linear equations presented here offers a valuable opportunity to explore fundamental concepts in linear algebra. These concepts include matrix operations, solving linear systems, and the geometric representation of solutions. Understanding these principles is crucial in various fields, including engineering, computer science, and economics, where linear systems are frequently encountered in modeling real-world problems. The geometric interpretation is particularly insightful as it bridges the gap between abstract algebraic equations and their visual counterparts, enhancing our intuition and understanding. This approach allows us to see how the solution to a system of equations corresponds to the point where the planes defined by the equations intersect in 3D space. By combining analytical methods with geometric visualization, we gain a comprehensive grasp of the solution's nature and its significance within the broader context of linear algebra.
System Definition
Let's define the system under consideration. We have the matrix A, the variable vector X, and the constant vector b as follows:
A = \begin{bmatrix}
1 & 2 & 2 \\
0 & 3 & 4 \\
0 & 0 & 2
\end{bmatrix}
X = \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
b = \begin{bmatrix}
5 \\
11 \\
6
\end{bmatrix}
The equation AX = b represents a system of three linear equations with three unknowns (x, y, and z). This system can be explicitly written as:
- x + 2y + 2z = 5
- 3y + 4z = 11
- 2z = 6
The matrix A is a crucial component of this system, as it encapsulates the coefficients of the variables in the equations. In this case, matrix A is an upper triangular matrix, which simplifies the process of solving the system. Upper triangular matrices have the property that all entries below the main diagonal are zero. This structure allows us to use a straightforward method called back-substitution to efficiently find the solution. The vector X represents the unknowns we aim to determine, and its components (x, y, z) are the values that satisfy all three equations simultaneously. The vector b on the right-hand side of the equation represents the constant terms, and it is essential for defining the specific solution we are seeking. Each equation in the system corresponds to a plane in three-dimensional space, and the solution (x, y, z) represents the point where all three planes intersect. Understanding these components and their roles is fundamental to solving the system and interpreting the solution geometrically.
Solving the System
To solve the system AX = b, we can use the method of back-substitution. This method is particularly effective for systems where the coefficient matrix is in upper triangular form, as is the case with matrix A.
From the third equation, 2z = 6, we can directly solve for z:
z = 6 / 2 = 3
Now that we have the value of z, we can substitute it into the second equation, 3y + 4z = 11, to solve for y:
3y + 4(3) = 11 3y + 12 = 11 3y = -1 y = -1/3
Finally, we substitute the values of y and z into the first equation, x + 2y + 2z = 5, to solve for x:
x + 2(-1/3) + 2(3) = 5 x - 2/3 + 6 = 5 x = 5 + 2/3 - 6 x = -1 + 2/3 x = -1/3
Therefore, the solution to the system is X = [-1/3, -1/3, 3]. The back-substitution method is a powerful technique for solving systems of linear equations, especially when the coefficient matrix has a triangular form. This method involves solving for one variable at a time, starting with the last equation and working backwards. The simplicity and efficiency of back-substitution make it a preferred approach for such systems. In our case, the upper triangular nature of matrix A allowed us to quickly determine the values of z, then y, and finally x. This step-by-step process highlights the elegance of linear algebra in simplifying complex problems. The solution vector X = [-1/3, -1/3, 3] represents the unique set of values for the variables (x, y, z) that satisfy all three equations in the system simultaneously. This solution is a specific point in three-dimensional space, which, as we will discuss in the next section, corresponds to the intersection of the planes represented by the equations.
Geometric Interpretation
Each equation in the system represents a plane in three-dimensional space. The solution to the system, X = [-1/3, -1/3, 3], corresponds to the point where all three planes intersect.
- The equation x + 2y + 2z = 5 represents a plane in 3D space.
- The equation 3y + 4z = 11 represents another plane in 3D space.
- The equation 2z = 6 represents yet another plane in 3D space.
The geometric interpretation of a system of linear equations is a fundamental concept in linear algebra, providing a visual understanding of the solutions. Each linear equation in a three-variable system can be visualized as a plane in three-dimensional space. The solutions to the system correspond to the points where these planes intersect. In our specific case, we have three equations, each representing a plane. The intersection of these three planes determines the solution set of the system. When the system has a unique solution, as in our case, the three planes intersect at a single point. This point, represented by the vector X = [-1/3, -1/3, 3], is the only point that lies on all three planes simultaneously, thus satisfying all three equations. If the planes do not intersect at a single point, the system may have infinitely many solutions (if the planes intersect along a line) or no solutions (if the planes are parallel or do not share a common intersection). The geometric perspective not only aids in visualizing the algebraic solutions but also provides insights into the nature of the system itself, such as its consistency and the uniqueness of solutions.
Conclusion
We have successfully solved the system AX = b using back-substitution, obtaining the solution X = [-1/3, -1/3, 3]. Geometrically, this solution represents the point of intersection of the three planes defined by the equations in the system.
Understanding both the algebraic and geometric aspects of solving linear systems is crucial in various applications. The process of solving the system AX = b involved several key steps, from setting up the equations to applying back-substitution and arriving at the solution vector X. This process highlights the importance of algebraic techniques in finding precise numerical solutions. However, the geometric interpretation adds another layer of understanding, allowing us to visualize the equations as planes in space and the solution as their intersection point. This dual perspective, combining algebraic precision with geometric intuition, is invaluable in problem-solving and modeling in various fields. The ability to translate between algebraic equations and geometric representations enhances our ability to understand and solve complex problems. Moreover, this approach provides a solid foundation for tackling more advanced topics in linear algebra and its applications, reinforcing the interconnectedness of mathematical concepts.
