Problem 57 - Intermediate Algebra

Question

Use Cramer's rule to solve each system of equations. If a system is inconsistent or if the equations are dependent, so indicate. $$ \left\\{\begin{array}{l} y=\frac{-2 x+1}{3} \\ 3 x-2 y=8 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y) = (2, -1)\).

Step 1: Reorganize the Equations

First, rewrite both equations in standard form to better apply Cramer's rule.The first equation given is \( y = \frac{-2x + 1}{3} \), which we can rearrange to: \( -2x - 3y = -1 \).The second equation is already in standard form: \( 3x - 2y = 8 \).Now, the system of equations becomes:\[-2x - 3y = -1\] \[3x - 2y = 8\]

Step 2: Set Up the Matrix for Cramer's Rule

For Cramer's rule, identify the matrix \( A \), determinant \( D \), and matrices for \( x \) and \( y \) values, \( D_x \) and \( D_y \).Matrix \( A \) is composed of the coefficients of \( x \) and \( y \):\( A = \begin{pmatrix} -2 & -3 \ 3 & -2 \end{pmatrix} \).

Step 3: Calculate Determinant D of Matrix A

The determinant \( D \) of matrix \( A \) is calculated as follows:\[ D = (-2)(-2) - (3)(-3) = 4 + 9 = 13 \].Since \( D eq 0 \), the system has a unique solution.

Step 4: Create Matrices Dx and Dy

Replace the respective column with the constants for calculating \( D_x \) and \( D_y \).For \( D_x \), replace the first column with constants:\( D_x = \begin{pmatrix} -1 & -3 \ 8 & -2 \end{pmatrix} \).For \( D_y \), replace the second column with constants:\( D_y = \begin{pmatrix} -2 & -1 \ 3 & 8 \end{pmatrix} \).

Step 5: Calculate Determinants Dx and Dy

Calculate \( D_x \) and \( D_y \) using determinants:\[ D_x = (-1)(-2) - (-3)(8) = 2 + 24 = 26 \].\[ D_y = (-2)(8) - (-1)(3) = -16 + 3 = -13 \].

Step 6: Calculate Variables x and y Using Cramer's Rule

Use the determinants to find \( x \) and \( y \).\[ x = \frac{D_x}{D} = \frac{26}{13} = 2 \].\[ y = \frac{D_y}{D} = \frac{-13}{13} = -1 \].

Step 7: Verify the Solution

Substitute \( x = 2 \) and \( y = -1 \) back into one of the original equations to verify.For \( y = \frac{-2x + 1}{3} \):\[ -1 = \frac{-2(2) + 1}{3} = \frac{-4 + 1}{3} = -1 \].The values satisfy the original equation, confirming the solution is correct.

Key Concepts

Systems of Equations

A system of equations consists of two or more equations with the same set of variables. In simpler terms, it's like having multiple recipes that use the same ingredients, and you are trying to find the right combination of those ingredients, represented by the variables, to satisfy all recipes simultaneously. In the problem discussed, we have two equations with two variables, namely, \(x\) and \(y\). These are known as linear equations because they are straight-line equations when graphically represented.
Two key things happen when solving systems of equations:

• Finding a solution that fits all equations. This means finding exact values for \(x\) and \(y\) that work in both equations.
• Determining whether the equations are consistent. Consistent systems have at least one set of solutions, while inconsistent ones do not.

For solving, multiple methods exist such as substitution, elimination, or using matrix techniques like Cramer's Rule.

Determinants

In the realm of matrix algebra, a determinant provides a single number that summarizes certain properties of matrices, often used in solving systems of equations.
For our case, the determinant helps to characterize the system and find out if a unique solution exists. The determinant of a 2x2 matrix:\[A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\]is calculated as:\[D = ad - bc.\]If \(D eq 0\), the system has a unique solution; this means there's exactly one combination of \(x\) and \(y\) values that satisfy both equations.
Conversely, if \(D = 0\), the system might have no solution or infinite solutions, indicating dependent or inconsistent equations.
Determinants play a crucial role in determining the viability of finding solutions via Cramer's Rule as demonstrated in solving our given exercise.

Matrix Algebra

Matrix algebra is an essential part of linear algebra and is the toolbox for manipulating systems of equations. A matrix is a structured array of numbers, often used to represent systems of linear equations compactly. Each row represents an equation, while each column represents a variable.
In the context of Cramer's Rule, we extract key matrices from the system:

• Matrix \( A \), which contains coefficients of variables from the system.
• Matrices \( D_x \) and \( D_y \), formed by replacing respective columns in \( A \) with the constants from the system.

The elegance of using matrices lies in their ability to handle complex calculations thus simplifying the process of solving systems of equations. With matrix algebra, we can visualize and tackle multiple equations in a streamlined manner, as shown in our exercise solution.

Unique Solutions

A system of equations with unique solutions has exactly one solution—one set of values for the variables that satisfy all equations. In the exercise, the determinant \( D \) is non-zero, which directly tells us the system will have a unique solution.
The process involves:

• Calculating determinants \( D, D_x, \) and \( D_y \).
• Determining the values of \( x \) and \( y \) using the formulas: \[x = \frac{D_x}{D}\, y = \frac{D_y}{D}\ \]

A unique solution means the equations intersect at a single point on a graph, and in practical terms, this translates to having one specific combination of variables that satisfies both equations perfectly.
This confirms the system builds onto a well-defined solution if all calculations align correctly, as illustrated in solving our given system.