Problem 32 - Intermediate Algebra

Question

Use Cramer's rule to find the solution set for each of the following systems. (Objective 2) $$ \left(\begin{array}{l} x-2 y=-1 \\ x=-6 y+5 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution set is \(\left(\frac{1}{2}, \frac{3}{4}\right)\).

Step 1: Write the system in standard form

Rewrite the given system of equations in standard form:\[\begin{align*}x - 2y &= -1, \x + 6y &= 5.\end{align*}\]

Step 2: Express in matrix form

The system of equations can be represented in matrix form as:\[A\mathbf{x} = \mathbf{b},\]where \(A = \begin{pmatrix} 1 & -2 \ 1 & 6 \end{pmatrix}\), \(\mathbf{x} = \begin{pmatrix} x \ y \end{pmatrix}\), and \(\mathbf{b} = \begin{pmatrix} -1 \ 5 \end{pmatrix}\).

Step 3: Calculate the determinant of the coefficient matrix

Compute the determinant \(\Delta\) of matrix \(A\):\[\Delta = \begin{vmatrix} 1 & -2 \ 1 & 6 \end{vmatrix} = (1)(6) - (1)(-2) = 6 + 2 = 8.\]

Step 4: Calculate determinants for Cramer's rule

Replace the first column of \(A\) with \(\mathbf{b}\) to compute \(\Delta_x\):\[\Delta_x = \begin{vmatrix} -1 & -2 \ 5 & 6 \end{vmatrix} = (-1)(6) - (5)(-2) = -6 + 10 = 4.\]Replace the second column of \(A\) with \(\mathbf{b}\) to compute \(\Delta_y\):\[\Delta_y = \begin{vmatrix} 1 & -1 \ 1 & 5 \end{vmatrix} = (1)(5) - (-1)(1) = 5 + 1 = 6.\]

Step 5: Apply Cramer's rule

Using Cramer's rule, calculate the solutions for \(x\) and \(y\):\[x = \frac{\Delta_x}{\Delta} = \frac{4}{8} = \frac{1}{2},\]\[y = \frac{\Delta_y}{\Delta} = \frac{6}{8} = \frac{3}{4}.\]

Step 6: Write the solution set

The solution set for the system of equations is:\[\left(\frac{1}{2}, \frac{3}{4}\right).\]

Key Concepts

Understanding Systems of Equations

A system of equations consists of two or more equations that share the same set of variables, typically representing a situation with multiple conditions that need to be satisfied at the same time. Solving a system of equations involves finding values for the variables that make all the equations true.
In this exercise, we have the system of equations with two equations:

• Equation 1: \(x - 2y = -1\)
• Equation 2: \(x = -6y + 5\)

These equations can represent anything from intersection points of lines to solutions in real-world scenarios like physics or economics. The goal is to find the intersection point (if it exists) that satisfies both equations, resulting in solutions for \(x\) and \(y\). By converting these equations into matrix form, we create a structured way to apply mathematical techniques like Cramer's rule.

The Role of Matrix Determinants

Matrix determinants play a crucial role in providing solutions to systems of linear equations, especially when using techniques like Cramer's Rule. A determinant is a special number that can be calculated from a square matrix, giving us important information about the matrix.
For our matrix \(A\), which is derived from the coefficients of the system of equations:

• Matrix \(A\) is \( \begin{pmatrix} 1 & -2 \ 1 & 6 \end{pmatrix} \)
• The determinant \(\Delta\) of \(A\) is calculated as \((1)(6) - (1)(-2) = 8\)

A non-zero determinant, like 8 in this case, indicates that the system of equations has a unique solution.
When using Cramer's Rule, determinants for modified matrices \(\Delta_x\) and \(\Delta_y\) are also calculated by replacing columns of the original matrix. This helps determine the solutions for the variables.

Application of Linear Algebra in Solving Equations

Linear algebra provides powerful tools for manipulating and solving systems of equations, with Cramer's Rule being one of these techniques. In our solution:

• We use matrices to represent the system of equations compactly.
• Determinants of matrices are used to apply Cramer's Rule.

With Cramer's Rule, we solve for each variable independently by using determinants:

• \(x = \frac{\Delta_x}{\Delta} = \frac{4}{8} = \frac{1}{2}\)
• \(y = \frac{\Delta_y}{\Delta} = \frac{6}{8} = \frac{3}{4}\)

These calculations illustrate linear algebra's ability to break down complex problems into manageable computational steps. This structured approach not only helps find precise solutions but also deepens our understanding of the system of equations being analyzed.