- Represent the linear system as the matrix equation \(AX = B\), with \(A = \begin{pmatrix} 2 & -3 \ 1 & 2 \end{pmatrix}\), \(X = \begin{pmatrix} x \ y \end{pmatrix}\), and \(B = \begin{pmatrix} 19 \ -1 \end{pmatrix}\).
- Calculate the inverse of matrix \(A\) as \(A^{-1} = \frac{1}{7} \begin{pmatrix} 2 & 3 \ -1 & 2 \end{pmatrix}\).
- Multiply \(A^{-1}\) by the constant vector \(B\) to solve for the variable vector \(X = A^{-1}B\).
- Conclude that the solution is \(\boxed{x=5, y=-3}\).
Explanation
- Matrix Representation
The system of equations is given by:
\[ \begin{array}{c} 2x - 3y = 19 \ x + 2y = -1 \end{array} \]
To solve this using the inverse matrix method, we first express the system as a matrix equation \(AX = B\), where:
\[ A = \begin{pmatrix} 2 & -3 \ 1 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \ y \end{pmatrix}, \quad B = \begin{pmatrix} 19 \ -1 \end{pmatrix} \]
Our goal is to find \(X\) by computing \(X = A^{-1}B\).
- Calculating the Inverse Matrix
First, we find the determinant of the coefficient matrix \(A\):
\[ \det(A) = (2)(2) - (-3)(1) = 4 + 3 = 7. \]
The inverse of a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by \(\frac{1}{\det(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}\). Applying this to \(A\):
\[ A^{-1} = \frac{1}{7} \begin{pmatrix} 2 & 3 \ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{2}{7} & \frac{3}{7} \ -\frac{1}{7} & \frac{2}{7} \end{pmatrix}. \]
- Solving the System
To find \(X\), we multiply the inverse matrix \(A^{-1}\) by the column vector \(B\):
\[ X = A^{-1}B = \frac{1}{7} \begin{pmatrix} 2 & 3 \ -1 & 2 \end{pmatrix} \begin{pmatrix} 19 \ -1 \end{pmatrix} \]
Performing the matrix multiplication:
\[ X = \frac{1}{7} \begin{pmatrix} 2(19) + 3(-1) \ -1(19) + 2(-1) \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 38 - 3 \ -19 - 2 \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 35 \ -21 \end{pmatrix} \]
Simplifying the resulting vector, we obtain:
\[ \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 5 \ -3 \end{pmatrix} \]
Thus, the solution to the system is \(x = 5\) and \(y = -3\).
Examples
In logistics, the inverse matrix method is used to balance supply chains. If a company knows how different routes contribute to the total delivery time for two types of products, they can set up a system of linear equations where \(x\) and \(y\) are the number of deliveries of each type. Solving this system using matrices allows managers to quickly determine the optimal number of deliveries for any given total time target, ensuring efficient resource allocation across the network.