Problem 29 - College Algebra

Question

For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{array}{c} 4 x+3 y=23 \\ 2 x-y=-1 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = 2\) and \(y = 5\).

Step 1: Write the System in Standard Form

Ensure the system is in the standard form of linear equations. The given system is already in standard form: \(4x + 3y = 23\) and \(2x - y = -1\).

Step 2: Construct the Coefficient Matrix

Identify and write down the coefficient matrix \(A\) from the system of equations. For \(4x + 3y = 23\) and \(2x - y = -1\), the coefficient matrix is \(A = \begin{bmatrix} 4 & 3 \ 2 & -1 \end{bmatrix}\).

Step 3: Compute the Determinant of the Coefficient Matrix

Calculate the determinant of matrix \(A\). If \(A = \begin{bmatrix} 4 & 3 \ 2 & -1 \end{bmatrix}\), then \(\det(A) = 4(-1) - 3(2) = -4 - 6 = -10\).

Step 4: Calculate Determinant for x (Replace x-coefficients)

Replace the x-coefficients in \(A\) with the constants from the equations to create matrix \(A_x\). So, \(A_x = \begin{bmatrix} 23 & 3 \ -1 & -1 \end{bmatrix}\). Now, calculate \(\det(A_x) = 23(-1) - 3(-1) = -23 + 3 = -20\).

Step 5: Calculate Determinant for y (Replace y-coefficients)

Replace the y-coefficients in \(A\) with the constants from the equations to create matrix \(A_y\). So, \(A_y = \begin{bmatrix} 4 & 23 \ 2 & -1 \end{bmatrix}\). Now, calculate \(\det(A_y) = 4(-1) - 23(2) = -4 - 46 = -50\).

Step 6: Apply Cramer's Rule to Solve for x and y

Using Cramer's Rule, solve for \(x\) and \(y\). The solution is given by \(x = \frac{\det(A_x)}{\det(A)} = \frac{-20}{-10} = 2\) and \(y = \frac{\det(A_y)}{\det(A)} = \frac{-50}{-10} = 5\).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations with multiple variables.
The goal is to find the values of these variables that satisfy all the equations simultaneously.
For example, in the system given:

• Equation 1: \(4x + 3y = 23\)
• Equation 2: \(2x - y = -1\)

This means you are trying to find values of \(x\) and \(y\) that make both equations true.
Such systems can have a single solution, no solution, or infinitely many solutions depending on the equations' characteristics.
Methods like substitution, elimination, and Cramer's Rule are popular for solving these systems.

Determinant

The determinant is a unique number that can be calculated from a square matrix.
For a 2x2 matrix, the determinant helps evaluate the matrix's properties, like singularity.
Given matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), its determinant is calculated as:

• \(\det(A) = ad - bc\)

In systems of linear equations, the determinant helps determine whether a unique solution exists.
If the determinant is zero, the system might have no solutions or an infinite number of solutions.
In contrast, a non-zero determinant indicates exactly one solution.

Coefficient Matrix

The coefficient matrix is derived from a system of linear equations by extracting the coefficients of the variables.
This matrix plays a crucial role in methods like Cramer's Rule.
For the given system:

• \(4x + 3y = 23\)
• \(2x - y = -1\)

The coefficient matrix \(A\) is:

• \(A = \begin{bmatrix} 4 & 3 \ 2 & -1 \end{bmatrix}\)

This matrix is used to calculate the determinant and to alter, forming new matrices when applying Cramer's Rule.
The coefficients provide a compact way to represent the linear aspects of the system.

Solving Equations

Solving equations involves finding the values of variables that satisfy the entire set of equations.
In the context of Cramer's Rule, this process includes several steps:

• Identifying the coefficient matrix from the system.
• Calculating the determinant of the coefficient matrix.
• Formulating new matrices for each variable with constants replacing a column.
• Computing determinants of these new matrices.
• Applying Cramer's Rule: \(x = \frac{\det(A_x)}{\det(A)}\) and \(y = \frac{\det(A_y)}{\det(A)}\).

This method, although systematic, is primarily useful for small systems, as it becomes complex for larger sizes.
Nevertheless, it ensures an exact solution whenever the determinant of the coefficient matrix is non-zero.