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Solve system using Cramer's rule. \(\frac{1}{2} x+\frac{2}{3} y=4\) \(\frac{3}{4} x+\frac{1}{3} y=-2\)

Short Answer

Expert verified
x = -8, y = 12

Step by step solution

01

- Write the system in matrix form

Express the given system of equations in matrix form: \[\begin{align*}\frac{1}{2} x + \frac{2}{3} y &= 4 \ \frac{3}{4} x + \frac{1}{3} y &= -2 \end{align*}\] This can be written as \[ AX = B \], where:\[A = \begin{pmatrix} \frac{1}{2} & \frac{2}{3} \ \frac{3}{4} & \frac{1}{3} \end{pmatrix}, X = \begin{pmatrix} x \ y \end{pmatrix}, B = \begin{pmatrix} 4 \ -2 \end{pmatrix} \]
02

- Calculate the determinant of matrix A (\(\Delta\))

The determinant of matrix A is calculated as:\[\Delta = \begin{vmatrix} \frac{1}{2} & \frac{2}{3} \ \frac{3}{4} & \frac{1}{3} \end{vmatrix} = \left( \frac{1}{2} \cdot \frac{1}{3} \right) - \left( \frac{2}{3} \cdot \frac{3}{4} \right) = \frac{1}{6} - \frac{1}{2} = \frac{1}{6} - \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3}\]
03

- Form matrix A_x and calculate its determinant (\(\Delta_x\))

To form the matrix \(A_x\), replace the first column of matrix A with the column matrix B:\[A_x = \begin{pmatrix} 4 & \frac{2}{3} \ -2 & \frac{1}{3} \end{pmatrix} \]The determinant \(\Delta_x\) is:\[\Delta_x = \begin{vmatrix} 4 & \frac{2}{3} \ -2 & \frac{1}{3} \end{vmatrix} = \left( 4 \cdot \frac{1}{3} \right) - \left( \frac{2}{3} \cdot -2 \right) = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}\]
04

- Form matrix A_y and calculate its determinant (\(\Delta_y\))

To form the matrix \(A_y\), replace the second column of matrix A with the column matrix B:\[A_y = \begin{pmatrix} \frac{1}{2} & 4 \ \frac{3}{4} & -2 \end{pmatrix} \]The determinant \(\Delta_y\) is:\[\Delta_y = \begin{vmatrix} \frac{1}{2} & 4 \ \frac{3}{4} & -2 \end{vmatrix} = \left( \frac{1}{2} \cdot -2 \right) - \left( 4 \cdot \frac{3}{4} \right) = -1 - 3 = -4\]
05

- Find the solution for x and y

Using Cramer's Rule, the values of x and y are determined by:\[x = \frac{\Delta_x}{\Delta} = \frac{\frac{8}{3}}{-\frac{1}{3}} = -8\]\[y = \frac{\Delta_y}{\Delta} = \frac{-4}{-\frac{1}{3}} = 12\]So, the solutions are:\[ x = -8 \ y = 12 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Linear Equations
A system of linear equations consists of two or more linear equations involving the same set of variables. In our exercise, we have two linear equations with variables x and y. The goal is to find the values of x and y that satisfy both equations simultaneously.
Here's the given system:
$$ \frac{1}{2} x + \frac{2}{3} y = 4 $$
$$ \frac{3}{4} x + \frac{1}{3} y = -2 $$
The solution to a system of linear equations can be graphically interpreted as the point(s) where the corresponding lines intersect. However, algebraic methods such as substitution, elimination, or matrix methods (including Cramer's rule) are often used for more accurate results. In this exercise, Cramer's rule helps by transforming the system into matrices and determinants, making it efficient to solve.
Determinants
Determinants are scalar values that can be computed from a square matrix. They provide valuable information about matrices, including invertibility and solutions to systems of linear equations. For a 2x2 matrix:
$$ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} $$
The determinant (denoted as det(A) or |A|) is calculated as:
$$ |A| = ad - bc $$
In our exercise, we found the determinant of matrix A:
$$ A = \begin{pmatrix} \frac{1}{2} & \frac{2}{3} \ \frac{3}{4} & \frac{1}{3} \end{pmatrix} $$
The determinant of A is calculated as:
$$ |A| = (\frac{1}{2} \cdot \frac{1}{3}) - (\frac{2}{3} \cdot \frac{3}{4}) = \frac{1}{6} - \frac{2}{3} = -\frac{1}{3} $$
Determinants play a crucial role in using Cramer's rule, as they help in finding the unique solution for variables if the determinant is non-zero.
Matrix Algebra
Matrix algebra involves mathematical operations with matrices, such as addition, subtraction, multiplication, and finding determinants. A matrix is a rectangular array of numbers arranged in rows and columns. Matrices are extensively used in solving systems of linear equations.
In our exercise, we use matrix notation to represent the system of equations:
$$ AX = B $$
Here, A, X, and B are matrices defined as follows:
$$ A = \begin{pmatrix} \frac{1}{2} & \frac{2}{3} \ \frac{3}{4} & \frac{1}{3} \end{pmatrix} $$
$$ X = \begin{pmatrix} x \ y \end{pmatrix} $$
$$ B = \begin{pmatrix} 4 \ - 2 \end{pmatrix} $$
By forming augmented matrices for x and y (A_x and A_y) and calculating their determinants, we can use Cramer's rule to solve for the variables:
$$ x = \frac{\Delta_x}{\Delta} $$
$$ y = \frac{\Delta_y}{\Delta} $$
Understanding the matrix form of systems of equations greatly simplifies complex algebraic processes and allows for efficient solutions.

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