Solve By Matrix Method: $ \frac{3x}{2} + 2y = 1, \quad \frac{y}{3} = 1 + \frac{x}{3} }$Ans { X = \frac{-10 {7} $}$
Introduction
In this article, we will explore the matrix method for solving a system of linear equations. This method is a powerful tool for solving systems of linear equations and is widely used in various fields such as engineering, economics, and computer science. We will use the given system of linear equations to demonstrate the matrix method and show how it can be used to find the solution.
The Matrix Method
The matrix method involves representing the system of linear equations as a matrix equation. This is done by creating a matrix of coefficients, a matrix of variables, and a matrix of constants. The matrix of coefficients is created by taking the coefficients of the variables in the system of linear equations. The matrix of variables is created by taking the variables in the system of linear equations. The matrix of constants is created by taking the constants in the system of linear equations.
Creating the Matrix of Coefficients
The matrix of coefficients is created by taking the coefficients of the variables in the system of linear equations. In this case, the system of linear equations is:
{ \frac{3x}{2} + 2y = 1, \quad \frac{y}{3} = 1 + \frac{x}{3} \}
The matrix of coefficients is:
{ \begin{bmatrix} \frac{3}{2} & 2 \\ 1 & -\frac{1}{3} \end{bmatrix} \}
Creating the Matrix of Variables
The matrix of variables is created by taking the variables in the system of linear equations. In this case, the system of linear equations is:
{ \frac{3x}{2} + 2y = 1, \quad \frac{y}{3} = 1 + \frac{x}{3} \}
The matrix of variables is:
{ \begin{bmatrix} x \\ y \end{bmatrix} \}
Creating the Matrix of Constants
The matrix of constants is created by taking the constants in the system of linear equations. In this case, the system of linear equations is:
{ \frac{3x}{2} + 2y = 1, \quad \frac{y}{3} = 1 + \frac{x}{3} \}
The matrix of constants is:
{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \}
Representing the System of Linear Equations as a Matrix Equation
The system of linear equations can be represented as a matrix equation by multiplying the matrix of coefficients by the matrix of variables and setting it equal to the matrix of constants. This is done by using the following equation:
{ \begin{bmatrix} \frac{3}{2} & 2 \\ 1 & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \}
Solving the Matrix Equation
To solve the matrix equation, we need to find the inverse of the matrix of coefficients. The inverse of a matrix is a matrix that, when multiplied by the original matrix, results in the identity matrix. The identity matrix is a matrix with ones on the main diagonal and zeros elsewhere.
Finding the Inverse of the Matrix of Coefficients
The inverse of the matrix of coefficients is found by using the following formula:
{ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix} \}
where is the determinant of the matrix of coefficients.
The determinant of the matrix of coefficients is:
{ \det(A) = \frac{3}{2} \cdot \left(-\frac{1}{3}\right) - 2 \cdot 1 = -\frac{1}{2} - 2 = -\frac{5}{2} \}
The inverse of the matrix of coefficients is:
{ A^{-1} = \frac{1}{-\frac{5}{2}} \begin{bmatrix} -\frac{1}{3} & -2 \\ -1 & \frac{3}{2} \end{bmatrix} = \begin{bmatrix} \frac{2}{15} & \frac{4}{5} \\ \frac{2}{5} & -\frac{3}{5} \end{bmatrix} \}
Multiplying the Inverse of the Matrix of Coefficients by the Matrix of Constants
To find the solution to the matrix equation, we need to multiply the inverse of the matrix of coefficients by the matrix of constants. This is done by using the following equation:
{ \begin{bmatrix} \frac{2}{15} & \frac{4}{5} \\ \frac{2}{5} & -\frac{3}{5} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{15} + \frac{4}{5} \\ \frac{2}{5} - \frac{3}{5} \end{bmatrix} = \begin{bmatrix} \frac{2}{15} + \frac{12}{15} \\ \frac{2}{5} - \frac{3}{5} \end{bmatrix} = \begin{bmatrix} \frac{14}{15} \\ -\frac{1}{5} \end{bmatrix} \}
Finding the Solution to the System of Linear Equations
The solution to the system of linear equations is found by multiplying the inverse of the matrix of coefficients by the matrix of constants. This is done by using the following equation:
{ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{2}{15} & \frac{4}{5} \\ \frac{2}{5} & -\frac{3}{5} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{14}{15} \\ -\frac{1}{5} \end{bmatrix} \}
Conclusion
In this article, we have demonstrated the matrix method for solving a system of linear equations. We have shown how to create the matrix of coefficients, the matrix of variables, and the matrix of constants. We have also shown how to find the inverse of the matrix of coefficients and multiply it by the matrix of constants to find the solution to the system of linear equations. The matrix method is a powerful tool for solving systems of linear equations and is widely used in various fields.
Final Answer
The final answer is .
Introduction
In the previous article, we demonstrated the matrix method for solving a system of linear equations. We created the matrix of coefficients, the matrix of variables, and the matrix of constants, and then found the inverse of the matrix of coefficients and multiplied it by the matrix of constants to find the solution to the system of linear equations. In this article, we will answer some frequently asked questions about the matrix method for solving a system of linear equations.
Q&A
Q: What is the matrix method for solving a system of linear equations?
A: The matrix method for solving a system of linear equations involves representing the system of linear equations as a matrix equation. This is done by creating a matrix of coefficients, a matrix of variables, and a matrix of constants. The matrix of coefficients is created by taking the coefficients of the variables in the system of linear equations. The matrix of variables is created by taking the variables in the system of linear equations. The matrix of constants is created by taking the constants in the system of linear equations.
Q: How do I create the matrix of coefficients?
A: To create the matrix of coefficients, you need to take the coefficients of the variables in the system of linear equations. For example, if the system of linear equations is:
{ \frac{3x}{2} + 2y = 1, \quad \frac{y}{3} = 1 + \frac{x}{3} \}
The matrix of coefficients is:
{ \begin{bmatrix} \frac{3}{2} & 2 \\ 1 & -\frac{1}{3} \end{bmatrix} \}
Q: How do I create the matrix of variables?
A: To create the matrix of variables, you need to take the variables in the system of linear equations. For example, if the system of linear equations is:
{ \frac{3x}{2} + 2y = 1, \quad \frac{y}{3} = 1 + \frac{x}{3} \}
The matrix of variables is:
{ \begin{bmatrix} x \\ y \end{bmatrix} \}
Q: How do I create the matrix of constants?
A: To create the matrix of constants, you need to take the constants in the system of linear equations. For example, if the system of linear equations is:
{ \frac{3x}{2} + 2y = 1, \quad \frac{y}{3} = 1 + \frac{x}{3} \}
The matrix of constants is:
{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \}
Q: How do I find the inverse of the matrix of coefficients?
A: To find the inverse of the matrix of coefficients, you need to use the following formula:
{ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix} \}
where is the determinant of the matrix of coefficients.
Q: How do I multiply the inverse of the matrix of coefficients by the matrix of constants?
A: To multiply the inverse of the matrix of coefficients by the matrix of constants, you need to use the following equation:
{ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{2}{15} & \frac{4}{5} \\ \frac{2}{5} & -\frac{3}{5} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{14}{15} \\ -\frac{1}{5} \end{bmatrix} \}
Q: What is the solution to the system of linear equations?
A: The solution to the system of linear equations is:
{ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{-10}{7} \\ \frac{6}{7} \end{bmatrix} \}
Conclusion
In this article, we have answered some frequently asked questions about the matrix method for solving a system of linear equations. We have shown how to create the matrix of coefficients, the matrix of variables, and the matrix of constants, and then found the inverse of the matrix of coefficients and multiplied it by the matrix of constants to find the solution to the system of linear equations. The matrix method is a powerful tool for solving systems of linear equations and is widely used in various fields.
Final Answer
The final answer is .