Question

Use an inverse matrix to solve (if possible) the system of linear equations. $$ \left\\{\begin{array}{l} \frac{5}{6} x-y=-20 \\ \frac{4}{3} x-\frac{7}{2} y=-51 \end{array}\right. $$

Short answer

The solutions to the system of equations is approximately \( x = 14.43 \) and \( y = 15.22 \)

Step-by-step solution

Express the system as a matrix equation

We can express the given system of equations in matrix form as follows: \[ \ \frac{5}{6} & -1 \ \frac{4}{3} & -\frac{7}{2} \ \] \( X = \) \[ \ x \ y \ \] and \( B = \) \[ \ -20 \ -51 \ \]. So this can be expressed as \( AX = B \) where A is the matrix of coefficients, X is the matrix of variables and B is the matrix of constants.

Find the inverse of matrix A

Solving for the inverse of A, denoted as \( A^{-1} \), is done using the formula \( A^{-1} = \frac{1}{det(A)} adj(A) \), where det(A) is the determinant of A and adj(A) is the adjoint of A. The determinant of A is given by \( det(A) = a*d - b*c = \frac{5}{6} * -\frac{7}{2} - (-1) * \frac{4}{3} = \frac{-35}{12} - (-\frac{4}{3}) = -\frac{23}{12} \). The adjoint of A is the transposition of the matrix of cofactors. In this 2x2 matrix case it's: \[ \ -\frac{7}{2} & 1 \ -\frac{4}{3} & \frac{5}{6}\ \]. So \( A^{-1} = \frac{-1}{\frac{-23}{12}} * \[ \ -\frac{7}{2} & 1 \ -\frac{4}{3} & \frac{5}{6} \ \] = \[ \ \frac{14}{23} & -\frac{12}{23} \ \frac{8}{23} & -\frac{10}{23} \ \] \)

Solve for X

Finally, to find X which contains the variables x and y, multiply \( A^{-1} \) with B. Hence \( AX = B \) becomes \( A^{-1}AX = A^{-1}B \) which further simplifies to \( IX = A^{-1}B \) (where I is the identity matrix) and finally becomes \( X = A^{-1}B \). So, \( X = \[ \ \frac{14}{23} & -\frac{12}{23} \ \frac{8}{23} & -\frac{10}{23} \ \] * \[ \ -20 \ -51 \ \] =\[ \ \frac{14}{23} * -20 -\frac{12}{23} * -51 \ \frac{8}{23} * -20 -\frac{10}{23} * -51 \ \] = \[ \ -\frac{280}{23} +\frac{612}{23} \ -\frac{160}{23} +\frac{510}{23} \ \] = \[ \ \frac{332}{23} \ \frac{350}{23} \ \] = \[ \ 14.43 \ 15.22 \ \] \) So, x is approximately 14.43 and y is approximately 15.22.

Key concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. The goal is to find values for each variable that satisfy all of the equations simultaneously.
In our particular example, we have a system with two equations and two variables:

• \( \frac{5}{6} x - y = -20 \)
• \( \frac{4}{3} x - \frac{7}{2} y = -51 \)

Each equation represents a line in a coordinate system. Solving this system means finding the intersection point of these two lines, which corresponds to the values of \( x \) and \( y \) that satisfy both equations.
Various methods exist to solve such systems, including graphical methods, substitution, elimination, and using matrices.

Matrix Equation

A matrix equation represents a system of linear equations in terms of matrices. This form can make solving the system cleaner and structured, especially with larger sets of equations.
In matrix terms, the system \( AX = B \) is a compact way of writing the equations:

• \( A \) is the coefficient matrix, capturing the coefficients from the linear equations as its elements.
• \( X \) is the matrix of variables, constituting unknowns in the system.
• \( B \) is the constants matrix, containing the right-hand side values of the equations.

For our system:A is:\[\begin{bmatrix} \frac{5}{6} & -1 \\frac{4}{3} & -\frac{7}{2} \end{bmatrix}\]
X is:\[\begin{bmatrix} x \y \end{bmatrix}\]
B is:\[\begin{bmatrix} -20 \-51 \end{bmatrix} \]This matrix form allows us to apply powerful techniques from linear algebra, like finding inverse matrices, to solve for X efficiently.

Determinant of Matrix

The determinant of a matrix is a special number that can tell us a lot about the matrix itself. For a 2x2 matrix, it's calculated as \( ad-bc \) where \( a, b, c, \) and \( d \) are elements of the matrix:For matrix \( A \):\[A = \begin{bmatrix} \frac{5}{6} & -1 \\frac{4}{3} & -\frac{7}{2} \end{bmatrix}\]The determinant, \( det(A) \), is given by:\[det(A) = \frac{5}{6} \cdot \left(-\frac{7}{2}\right) - (-1) \cdot \frac{4}{3} = -\frac{35}{12} + \frac{4}{3} = -\frac{23}{12} \]
The determinant provides information about whether the matrix is invertible. If the determinant is zero, the matrix does not have an inverse, and thus the system of equations has either no solutions or infinitely many solutions.

Adjoint of Matrix

The adjoint of a matrix is the transpose of its cofactor matrix. For a 2x2 matrix, when written out, the adjoint looks as follows:Given matrix \( A \):\[A = \begin{bmatrix} \frac{5}{6} & -1 \\frac{4}{3} & -\frac{7}{2} \end{bmatrix}\]The adjoint, \( adj(A) \), would be:\[adj(A) = \begin{bmatrix} -\frac{7}{2} & 1 \-\frac{4}{3} & \frac{5}{6} \end{bmatrix}\]The adjoint is useful in calculating the inverse of a matrix. Specifically, for any matrix \( A \), the inverse is given by:\[A^{-1} = \frac{1}{det(A)} \cdot adj(A)\]This highlights the role of the adjoint in facilitating the process of finding the inverse of matrices used to solve linear system equations, provided the determinant is not zero.