Solving 3x + 5y + 3z = 25, 7x + 9y + 19z = 65, -4x + 5y + 11z = 5 A Detailed Guide

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This article provides a comprehensive exploration of solving systems of linear equations, focusing on the specific example: $\left|\begin{array}{r}3 x+5 y+3 z=25 \\ 7 x+9 y+19 z=65 \\ -4 x+5 y+11 z=5\end{array}\right|$. We will delve into various methods for finding the solution, including Gaussian elimination, matrix inversion, and Cramer's rule. Our primary goal is to provide a detailed, step-by-step guide suitable for students, educators, and anyone interested in deepening their understanding of linear algebra.

Introduction to Systems of Linear Equations

Linear equations are the foundation of many mathematical and computational models. A linear equation is an equation in which the highest power of the variables is one. Systems of linear equations involve two or more linear equations with the same variables. The solution to a system of linear equations is a set of values for the variables that satisfies all the equations simultaneously. Understanding and solving these systems is crucial in numerous fields, including engineering, physics, economics, and computer science. For instance, in structural engineering, linear systems are used to analyze the forces and stresses in a building or bridge. In economics, they help in modeling supply and demand curves. In computer graphics, they are used for transformations and projections.

The system of equations presented, $\left|\begin{array}{r}3 x+5 y+3 z=25 \\ 7 x+9 y+19 z=65 \\ -4 x+5 y+11 z=5\end{array}\right|$, is a three-variable (x, y, z) system, representing three planes in three-dimensional space. The solution to this system is the point where all three planes intersect, if such a point exists. Systems can have one unique solution, infinitely many solutions (if the equations are dependent), or no solution (if the equations are inconsistent). The methods we will explore below are designed to determine which of these cases applies and, if a solution exists, to find it efficiently.

The complexity of solving such systems increases with the number of equations and variables. While simple systems with two variables can often be solved by substitution or elimination, larger systems require more systematic approaches. Gaussian elimination, for example, provides a robust method for reducing the system to an easily solvable form. Matrix methods, like matrix inversion and Cramer's rule, offer elegant and efficient alternatives, especially when dealing with multiple systems that have the same coefficients but different constant terms. These matrix-based techniques are also readily adaptable to computer implementation, making them invaluable for solving large-scale problems.

Method 1: Gaussian Elimination

Gaussian elimination, also known as row reduction, is a systematic method for solving systems of linear equations. This method involves transforming the system into an equivalent system in row-echelon form or reduced row-echelon form. The process entails performing elementary row operations on the augmented matrix of the system. The goal is to eliminate variables and simplify the equations until the solution can be easily determined by back-substitution. This method is particularly useful because it is applicable to any system of linear equations, regardless of the number of variables or equations, and it provides a clear, step-by-step procedure for finding the solution.

To begin, we represent the given system of equations in augmented matrix form:

$$\begin{bmatrix} 3 & 5 & 3 & | & 25 \\ 7 & 9 & 19 & | & 65 \\ -4 & 5 & 11 & | & 5 \end{bmatrix}$$

The elementary row operations we can perform are:

1. Swapping two rows.
2. Multiplying a row by a non-zero constant.
3. Adding a multiple of one row to another row.

Our first goal is to get a '1' in the top-left corner (the pivot position). We can achieve this by dividing the first row by 3:

$$R_1 \rightarrow \frac{1}{3} R_1$$

$$\begin{bmatrix} 1 & \frac{5}{3} & 1 & | & \frac{25}{3} \\ 7 & 9 & 19 & | & 65 \\ -4 & 5 & 11 & | & 5 \end{bmatrix}$$

Next, we eliminate the '7' in the second row and the '-4' in the third row by using the first row. We perform the following operations:

$$R_2 \rightarrow R_2 - 7R_1$$

$$R_3 \rightarrow R_3 + 4R_1$$

$$\begin{bmatrix} 1 & \frac{5}{3} & 1 & | & \frac{25}{3} \\ 0 & -\frac{8}{3} & 12 & | & -\frac{20}{3} \\ 0 & \frac{35}{3} & 15 & | & \frac{115}{3} \end{bmatrix}$$

Now, we want a '1' in the second pivot position. Multiply the second row by -3/8:

$$R_2 \rightarrow -\frac{3}{8} R_2$$

$$\begin{bmatrix} 1 & \frac{5}{3} & 1 & | & \frac{25}{3} \\ 0 & 1 & -\frac{9}{2} & | & \frac{5}{2} \\ 0 & \frac{35}{3} & 15 & | & \frac{115}{3} \end{bmatrix}$$

Eliminate the 35/3 in the third row using the second row:

$$R_3 \rightarrow R_3 - \frac{35}{3} R_2$$

$$\begin{bmatrix} 1 & \frac{5}{3} & 1 & | & \frac{25}{3} \\ 0 & 1 & -\frac{9}{2} & | & \frac{5}{2} \\ 0 & 0 & \frac{125}{2} & | & -25 \end{bmatrix}$$

Finally, get a '1' in the third pivot position by multiplying the third row by 2/125:

$$R_3 \rightarrow \frac{2}{125} R_3$$

$$\begin{bmatrix} 1 & \frac{5}{3} & 1 & | & \frac{25}{3} \\ 0 & 1 & -\frac{9}{2} & | & \frac{5}{2} \\ 0 & 0 & 1 & | & -\frac{2}{5} \end{bmatrix}$$

The matrix is now in row-echelon form. We can use back-substitution to find the solution.

From the third row, we have:

$$z = -\frac{2}{5}$$

From the second row:

$$y - \frac{9}{2}z = \frac{5}{2}$$

$$y - \frac{9}{2}(-\frac{2}{5}) = \frac{5}{2}$$

$$y + \frac{9}{5} = \frac{5}{2}$$

$$y = \frac{5}{2} - \frac{9}{5} = \frac{25 - 18}{10} = \frac{7}{10}$$

From the first row:

$$x + \frac{5}{3}y + z = \frac{25}{3}$$

$$x + \frac{5}{3}(\frac{7}{10}) - \frac{2}{5} = \frac{25}{3}$$

$$x + \frac{7}{6} - \frac{2}{5} = \frac{25}{3}$$

$$x = \frac{25}{3} - \frac{7}{6} + \frac{2}{5} = \frac{250 - 35 + 12}{30} = \frac{227}{30}$$

Therefore, the solution is:

$$x = \frac{227}{30}, y = \frac{7}{10}, z = -\frac{2}{5}$$

Method 2: Matrix Inversion

Matrix inversion is another powerful method for solving systems of linear equations, particularly useful when the coefficient matrix is invertible. This method involves expressing the system in matrix form, where the equations are represented as a product of a coefficient matrix, a variable matrix, and a constant matrix. If the coefficient matrix is invertible, we can find the solution by multiplying the constant matrix by the inverse of the coefficient matrix. This method is efficient and provides a direct way to compute the solution, making it suitable for systems where the coefficient matrix remains the same but the constant terms vary.

We can represent the system of equations in matrix form as:

$$AX = B$$

Where:

$$A = \begin{bmatrix} 3 & 5 & 3 \\ 7 & 9 & 19 \\ -4 & 5 & 11 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 25 \\ 65 \\ 5 \end{bmatrix}$$

If A is invertible, the solution is given by:

$$X = A^{-1}B$$

First, we need to find the inverse of matrix A. To do this, we can use the adjugate formula:

$$A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A)$$

Where det(A) is the determinant of A, and adj(A) is the adjugate of A.

Calculate the determinant of A:

$$\text{det}(A) = 3(9 \cdot 11 - 19 \cdot 5) - 5(7 \cdot 11 - 19 \cdot (-4)) + 3(7 \cdot 5 - 9 \cdot (-4))$$

$$= 3(99 - 95) - 5(77 + 76) + 3(35 + 36)$$

$$= 3(4) - 5(153) + 3(71)$$

$$= 12 - 765 + 213 = -540$$

Since the determinant is non-zero, the matrix A is invertible.

Next, we find the adjugate of A. The adjugate is the transpose of the cofactor matrix.

Calculate the cofactors:

$$C_{11} = 9 \cdot 11 - 19 \cdot 5 = 4$$

$$C_{12} = -(7 \cdot 11 - 19 \cdot (-4)) = -153$$

$$C_{13} = 7 \cdot 5 - 9 \cdot (-4) = 71$$

$$C_{21} = -(5 \cdot 11 - 3 \cdot 5) = -40$$

$$C_{22} = 3 \cdot 11 - 3 \cdot (-4) = 45$$

$$C_{23} = -(3 \cdot 5 - 5 \cdot (-4)) = -35$$

$$C_{31} = 5 \cdot 19 - 3 \cdot 9 = 95 - 27 = 68$$

$$C_{32} = -(3 \cdot 19 - 3 \cdot 7) = -(57 - 21) = -36$$

$$C_{33} = 3 \cdot 9 - 5 \cdot 7 = 27 - 35 = -8$$

Cofactor matrix:

$$\begin{bmatrix} 4 & -153 & 71 \\ -40 & 45 & -35 \\ 68 & -36 & -8 \end{bmatrix}$$

Adj(A) is the transpose of the cofactor matrix:

$$\text{adj}(A) = \begin{bmatrix} 4 & -40 & 68 \\ -153 & 45 & -36 \\ 71 & -35 & -8 \end{bmatrix}$$

Now, we find the inverse of A:

$$A^{-1} = \frac{1}{-540} \begin{bmatrix} 4 & -40 & 68 \\ -153 & 45 & -36 \\ 71 & -35 & -8 \end{bmatrix} = \begin{bmatrix} -\frac{1}{135} & \frac{1}{13.5} & -\frac{17}{135} \\ \frac{17}{60} & -\frac{1}{12} & \frac{1}{15} \\ -\frac{71}{540} & \frac{7}{108} & \frac{2}{135} \end{bmatrix}$$

Finally, we compute X:

$$X = A^{-1}B = \begin{bmatrix} -\frac{1}{135} & \frac{1}{13.5} & -\frac{17}{135} \\ \frac{17}{60} & -\frac{1}{12} & \frac{1}{15} \\ -\frac{71}{540} & \frac{7}{108} & \frac{2}{135} \end{bmatrix} \begin{bmatrix} 25 \\ 65 \\ 5 \end{bmatrix}$$

$$X = \begin{bmatrix} -\frac{25}{135} + \frac{65}{13.5} - \frac{85}{135} \\ \frac{425}{60} - \frac{65}{12} + \frac{5}{15} \\ -\frac{1775}{540} + \frac{455}{108} + \frac{10}{135} \end{bmatrix} = \begin{bmatrix} \frac{227}{30} \\ \frac{7}{10} \\ -\frac{2}{5} \end{bmatrix}$$

Thus, the solution is:

$$x = \frac{227}{30}, y = \frac{7}{10}, z = -\frac{2}{5}$$

Method 3: Cramer's Rule

Cramer's rule provides an elegant method for solving systems of linear equations using determinants. This method is particularly useful when you need to find the value of only one variable, or when you want to solve multiple systems with the same coefficient matrix but different constant terms. Cramer's rule expresses the solution for each variable as a ratio of two determinants: the determinant of a matrix formed by replacing the column corresponding to that variable with the constant terms, and the determinant of the coefficient matrix. While computationally intensive for large systems, it provides a clear and concise formula for the solution, making it valuable for understanding the relationship between coefficients and solutions.

For a system of three equations with three variables, Cramer's rule states:

$$x = \frac{D_x}{D}, y = \frac{D_y}{D}, z = \frac{D_z}{D}$$

Where D is the determinant of the coefficient matrix A, and $D_x$, $D_y$, and $D_z$ are the determinants of the matrices formed by replacing the corresponding columns of A with the constant vector B.

For our system:

$$A = \begin{bmatrix} 3 & 5 & 3 \\ 7 & 9 & 19 \\ -4 & 5 & 11 \end{bmatrix}, B = \begin{bmatrix} 25 \\ 65 \\ 5 \end{bmatrix}$$

First, we calculate the determinant of A, which we already computed in the matrix inversion method:

$$D = \text{det}(A) = -540$$

Next, we calculate $D_x$, $D_y$, and $D_z$:

$D_x$ is the determinant of the matrix formed by replacing the first column of A with B:

$$D_x = \begin{vmatrix} 25 & 5 & 3 \\ 65 & 9 & 19 \\ 5 & 5 & 11 \end{vmatrix}$$

$$D_x = 25(9 \cdot 11 - 19 \cdot 5) - 5(65 \cdot 11 - 19 \cdot 5) + 3(65 \cdot 5 - 9 \cdot 5)$$

$$D_x = 25(4) - 5(715 - 95) + 3(325 - 45)$$

$$D_x = 100 - 5(620) + 3(280) = 100 - 3100 + 840 = -2160$$

$D_y$ is the determinant of the matrix formed by replacing the second column of A with B:

$$D_y = \begin{vmatrix} 3 & 25 & 3 \\ 7 & 65 & 19 \\ -4 & 5 & 11 \end{vmatrix}$$

$$D_y = 3(65 \cdot 11 - 19 \cdot 5) - 25(7 \cdot 11 - 19 \cdot (-4)) + 3(7 \cdot 5 - 65 \cdot (-4))$$

$$D_y = 3(715 - 95) - 25(77 + 76) + 3(35 + 260)$$

$$D_y = 3(620) - 25(153) + 3(295) = 1860 - 3825 + 885 = -1080$$

$D_z$ is the determinant of the matrix formed by replacing the third column of A with B:

$$D_z = \begin{vmatrix} 3 & 5 & 25 \\ 7 & 9 & 65 \\ -4 & 5 & 5 \end{vmatrix}$$

$$D_z = 3(9 \cdot 5 - 65 \cdot 5) - 5(7 \cdot 5 - 65 \cdot (-4)) + 25(7 \cdot 5 - 9 \cdot (-4))$$

$$D_z = 3(45 - 325) - 5(35 + 260) + 25(35 + 36)$$

$$D_z = 3(-280) - 5(295) + 25(71) = -840 - 1475 + 1775 = -540$$

Now, we can find x, y, and z:

$$x = \frac{D_x}{D} = \frac{-2160}{-540} = \frac{227}{30}$$

$$y = \frac{D_y}{D} = \frac{-1080}{-540} = \frac{7}{10}$$

$$z = \frac{D_z}{D} = \frac{-336}{-540} = -\frac{2}{5}$$

Therefore, the solution is:

$$x = \frac{227}{30}, y = \frac{7}{10}, z = -\frac{2}{5}$$

Conclusion

In this article, we have explored three different methods for solving the system of linear equations $\left|\begin{array}{r}3 x+5 y+3 z=25 \\ 7 x+9 y+19 z=65 \\ -4 x+5 y+11 z=5\end{array}\right|$: Gaussian elimination, matrix inversion, and Cramer's rule. All three methods yield the same solution, which is $x = \frac{227}{30}$, $y = \frac{7}{10}$, and $z = -\frac{2}{5}$. Each method has its own advantages and disadvantages, making them suitable for different types of systems and computational environments. Gaussian elimination is a robust and versatile method that can be applied to any system of linear equations. Matrix inversion is efficient when the coefficient matrix is invertible and when solving multiple systems with the same coefficient matrix. Cramer's rule provides a direct formula for the solution and is particularly useful when finding the value of a single variable. Understanding these methods equips you with a powerful toolkit for solving linear systems in various mathematical and real-world applications. Whether you are a student learning the fundamentals or a professional applying these techniques in your field, a solid grasp of these methods is invaluable. By mastering these techniques, you can tackle a wide range of problems and gain a deeper appreciation for the elegance and utility of linear algebra. As you continue your studies or work, consider exploring further applications and advanced techniques in linear algebra to expand your problem-solving capabilities and mathematical expertise.