Problem 60 - A Graphical Approach to College Algebra

Question

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}&4 x-y=0\\\&2 x+3 y=14\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \) and \( y = 4 \).

Step 1: Write the System of Equations

The system of equations is given as:\[\begin{align*} 4x - y &= 0 \ 2x + 3y &= 14 \end{align*}\] This shows two equations with two variables \(x\) and \(y\).

Step 2: Express in Matrix Form

Convert the system into matrix form \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constant matrix:\[ A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 0 \ 14 \end{bmatrix} \]

Step 3: Calculate the Determinant of Matrix A

Find the determinant \( D \) of the matrix \( A \):\[ D = \begin{vmatrix} 4 & -1 \ 2 & 3 \end{vmatrix} = (4)(3) - (2)(-1) = 12 + 2 = 14 \]Since \( D eq 0 \), Cramer's Rule can be used.

Step 4: Formulate the Matrices for Cramer's Rule

To find \( x \) and \( y \), replace the respective columns of \( A \) with \( \mathbf{b} \):- For \( x \):\[ A_x = \begin{bmatrix} 0 & -1 \ 14 & 3 \end{bmatrix} \]- For \( y \):\[ A_y = \begin{bmatrix} 4 & 0 \ 2 & 14 \end{bmatrix} \]

Step 5: Calculate Determinant for Each Matrix

Find the determinants \( D_x \) and \( D_y \):- For \( A_x \):\[ D_x = \begin{vmatrix} 0 & -1 \ 14 & 3 \end{vmatrix} = (0)(3) - (-1)(14) = 14 \]- For \( A_y \):\[ D_y = \begin{vmatrix} 4 & 0 \ 2 & 14 \end{vmatrix} = (4)(14) - (0)(2) = 56 \]

Step 6: Apply Cramer's Rule to Solve for Variables

Use Cramer's rule to solve for \( x \) and \( y \):\[ x = \frac{D_x}{D} = \frac{14}{14} = 1 \]\[ y = \frac{D_y}{D} = \frac{56}{14} = 4 \]

Step 7: Verify the Solution

Plug \( x = 1 \) and \( y = 4 \) back into the original equations to verify:- First Equation: \( 4(1) - 4 = 0 \) (True)- Second Equation: \( 2(1) + 3(4) = 2 + 12 = 14 \) (True)

Key Concepts

System of Equations

To understand Cramer's rule, we first need to grasp the concept of a system of equations. A system of equations involves several equations working together to find the values of unknown variables. In the given exercise, we have two equations:

• \(4x - y = 0\)
• \(2x + 3y = 14\)

These equations share the same variables, \(x\) and \(y\). The goal is to find specific values for \(x\) and \(y\) that satisfy both equations at the same time. We achieve this by solving the equations concurrently, ensuring that the solution works for all equations in the system.
Such systems are common in real-world problems, where multiple conditions must be satisfied simultaneously. The systematic approach taken by Cramer's rule provides us a clear method to solve these systems using the properties of matrices.

Determinant Calculation

A crucial aspect of using Cramer's rule is calculating determinants. The determinant is a special number that can be computed from a square matrix, and it plays a significant role in linear algebra.
In our example, we have the coefficient matrix:\[A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\]To determine if we can apply Cramer's rule, we first calculate the determinant of this matrix:\[D = \begin{vmatrix} 4 & -1 \ 2 & 3 \end{vmatrix} = (4)(3) - (2)(-1) = 14\]Because the determinant \(D\) is not zero, Cramer's rule is applicable.

• The computation of determinants follows specific rules based on the arrangement of numbers in the matrix.
• If the determinant equals zero, the matrix is singular, meaning the equations cannot be solved using Cramer's rule. We would need to use another method, such as substitution or elimination.

The determinant helps assess the system's solvability, guiding us in choosing the right approach.

Matrix Algebra

Matrix algebra provides a structured method to handle systems of equations, particularly when using Cramer's rule.
Using matrices to represent systems simplifies complex calculations and makes the process systematic and straightforward.When addressing the given two-equation system:

• We expressed it in matrix form as\[ A \mathbf{x} = \mathbf{b},\]where \(A\) is the coefficient matrix, \(\mathbf{x}\) is the variable matrix, and \(\mathbf{b}\) is the constant matrix.
• Matrix algebra allows us to manipulate matrices and perform operations, such as finding determinants or solving for variables using rules like Cramer's.
• By swapping columns in the coefficient matrix with the constant matrix \(\mathbf{b}\), we create new matrices, \(A_x\) and \(A_y\), to solve for individual variables \(x\) and \(y\).

This method is efficient for solving linear equations as it organizes calculations and offers an elegant way to address complex arithmetic. As you get comfortable with matrix operations, solving large systems of equations becomes less daunting and more intuitive.