Problem 31 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{array}{l} 4 x-3 y=-3 \\ 2 x+6 y=-4 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = -1\) and \(y = -\frac{1}{3}\).

Step 1: Write the system in matrix form

First, we rewrite the system of linear equations in matrix form \(Ax = b\), where \(A\) is the coefficient matrix, \(x\) is the column of variables, and \(b\) is the column of constants.\[ A = \begin{bmatrix} 4 & -3 \ 2 & 6 \end{bmatrix}, \quad x = \begin{bmatrix} x \ y \end{bmatrix}, \quad b = \begin{bmatrix} -3 \ -4 \end{bmatrix} \]

Step 2: Calculate the determinant of A

To use Cramer's Rule, we first calculate the determinant of the matrix \(A\).\[ \text{det}(A) = \begin{vmatrix} 4 & -3 \ 2 & 6 \end{vmatrix} = (4)(6) - (-3)(2) = 24 + 6 = 30 \]

Step 3: Calculate determinants for x and y

Next, we calculate the determinants for the modified matrices. Replace the first column of \(A\) with \(b\) to find \(\text{det}(A_x)\) and replace the second column with \(b\) to find \(\text{det}(A_y)\). \[ A_x = \begin{bmatrix} -3 & -3 \ -4 & 6 \end{bmatrix}, \quad \text{det}(A_x) = (-3)(6) - (-3)(-4) = -18 + 12 = -30 \] \[ A_y = \begin{bmatrix} 4 & -3 \ 2 & -4 \end{bmatrix}, \quad \text{det}(A_y) = (4)(-4) - (-3)(2) = -16 + 6 = -10 \]

Step 4: Solve for x and y using Cramer's Rule

Cramer's Rule states that the solution for \(x\) and \(y\) is given by \(x = \frac{\text{det}(A_x)}{\text{det}(A)}\) and \(y = \frac{\text{det}(A_y)}{\text{det}(A)}\).\[ x = \frac{-30}{30} = -1 \]\[ y = \frac{-10}{30} = -\frac{1}{3} \]

Key Concepts

System of Linear Equations

A system of linear equations is a set of two or more equations involving the same set of variables. The goal is to find values for these variables that satisfy all of the equations simultaneously. For instance, in the given exercise, we have two equations with the variables \(x\) and \(y\):

• \(4x - 3y = -3\)
• \(2x + 6y = -4\)

Such systems can be solved using various methods, including graphing, substitution, elimination, and Cramer's Rule, which is efficient when dealing with equations represented in matrix form. Understanding these concepts can significantly enhance our ability to solve real-world problems that can be modeled with linear equations.

Determinant of a Matrix

The determinant of a matrix is a special value that can be computed from its elements. It is a useful scalar quantity in linear algebra and appears in many important calculations, including solving systems of linear equations using Cramer's Rule. For a 2x2 matrix \(A\):\[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \]The determinant, denoted as \(\text{det}(A)\), is calculated as:\[ \text{det}(A) = ad - bc \]In this exercise, the determinant for the coefficient matrix \(A\) was calculated as follows:\[ \text{det}(A) = (4)(6) - (-3)(2) = 24 + 6 = 30 \]Determining this value is crucial for analyzing whether a system of equations has a unique solution, infinitely many solutions, or no solution at all.

Matrix Form

When solving systems of linear equations using matrices, it is crucial to express the system in matrix form. The matrix form consists of three parts:

• The coefficient matrix \(A\), containing all the coefficients of the variables.
• The variable vector \(x\), containing the variables \(x\) and \(y\).
• The constant vector \(b\), containing the constant terms from each equation.

In the given example:\[ A = \begin{bmatrix} 4 & -3 \ 2 & 6 \end{bmatrix}, \quad x = \begin{bmatrix} x \ y \end{bmatrix}, \quad b = \begin{bmatrix} -3 \ -4 \end{bmatrix} \]This representation gives us the linear matrix equation \(Ax = b\). Converting systems into this form allows for the application of various solution techniques such as Cramer's Rule, as well as offering insights into the properties and possible solutions of the system.