Problem 60 - A Graphical Approach to Precalculus with Limits

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 in Matrix Form

The system of equations can be written in matrix form as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables, and \(B\) is the column matrix of constants. Here, \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\), \(X = \begin{bmatrix} x \ y \end{bmatrix}\), and \(B = \begin{bmatrix} 0 \ 14 \end{bmatrix}\).

Step 2: Calculate the Determinant of the Coefficient Matrix

The determinant of matrix \(A\), denoted as \(|A|\), is calculated as \(|A| = (4)(3) - (2)(-1) = 12 + 2 = 14\). Since \(|A| eq 0\), we can proceed with Cramer's Rule.

Step 3: Find the Determinant of Matrix for x

To find \(x\) using Cramer's Rule, replace the first column of \(A\) with \(B\) to form a new matrix \(A_x = \begin{bmatrix} 0 & -1 \ 14 & 3 \end{bmatrix}\). Calculate the determinant of \(A_x\): \(|A_x| = (0)(3) - (14)(-1) = 0 + 14 = 14\).

Step 4: Find the Determinant of Matrix for y

To find \(y\), replace the second column of \(A\) with \(B\) to form a new matrix \(A_y = \begin{bmatrix} 4 & 0 \ 2 & 14 \end{bmatrix}\). Calculate the determinant of \(A_y\): \(|A_y| = (4)(14) - (2)(0) = 56\).

Step 5: Solve for x and y Using Cramer's Rule

Using Cramer's Rule, solve for \(x\) and \(y\): \(x = \frac{|A_x|}{|A|} = \frac{14}{14} = 1\) and \(y = \frac{|A_y|}{|A|} = \frac{56}{14} = 4\).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations that share a set of variables. In the given problem, we have two equations: \(4x - y = 0\) and \(2x + 3y = 14\). Each equation describes a line on a coordinate plane, and solving the system means finding the point(s) where these lines intersect, representing the solution for these variables.

There are several methods to solve systems of equations:

• Substitution
• Elimination
• Matrix methods, such as Cramer's Rule and Gaussian Elimination

Here, we focus on Cramer's Rule, as it provides a straightforward way to solve a system of linear equations using determinants.

Determinant of a Matrix

The determinant of a matrix is a special number that you can calculate from its elements. It provides crucial information about the matrix, such as whether it is invertible. For a 2x2 matrix, which is what we are using in this example, the determinant can be calculated as follows: given the matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant of \(A\), denoted \(|A|\), is \(|A| = ad - bc\).

In our exercise, we calculate the determinant of the coefficient matrix \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\). The determinant is \( (4 \times 3) - (2 \times -1) = 12 + 2 = 14 \). Since \(|A| eq 0\), we can proceed with Cramer's Rule, signifying that the system has a unique solution.

Matrix Form

To solve a system using matrices, we write it in a special format called matrix form. In matrix form, the system \(4x - y = 0\) and \(2x + 3y = 14\) can be expressed using matrices as \(AX = B\), where:

• \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\) is the coefficient matrix representing the coefficients of the variables \(x\) and \(y\).
• \(X = \begin{bmatrix} x \ y \end{bmatrix}\) is the column matrix of variables.
• \(B = \begin{bmatrix} 0 \ 14 \end{bmatrix}\) is the column matrix of constants from the right-hand side of the equations.

Writing the system in matrix form simplifies applying matrix methods like Cramer's Rule to find the solution.

Solving Equations

With Cramer's Rule, solving the system of equations involves replacing the columns of the coefficient matrix with the constant matrix one at a time to form new matrices. Each new matrix helps us find the value of one variable.

Here's how it works:

• For \(x\), replace the first column of \(A\) with \(B\) to form \(A_x = \begin{bmatrix} 0 & -1 \ 14 & 3 \end{bmatrix}\). The determinant is \(|A_x| = 14\).
• For \(y\), replace the second column of \(A\) with \(B\) to form \(A_y = \begin{bmatrix} 4 & 0 \ 2 & 14 \end{bmatrix}\). The determinant is \(|A_y| = 56\).

Finally, apply Cramer's Rule:
- \(x = \frac{|A_x|}{|A|} = \frac{14}{14} = 1\)
- \(y = \frac{|A_y|}{|A|} = \frac{56}{14} = 4\)
Thus, the solution \(x = 1, y = 4\) is the point where the two lines represented by the equations intersect.