Problem 30 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{array}{l} 10 x-6 y=2 \\ -5 x+8 y=-1 \end{array} $$

Step-by-Step Solution

Answer

The solution is \( x = 0.44 \) and \( y = 0 \).

Step 1: Write the System in Matrix Form

First, express the system of equations as a matrix equation. The system given is: \( 10x - 6y = 2 \) \( -5x + 8y = -1 \). This can be written in matrix form as: \[ \begin{bmatrix} 10 & -6 \ -5 & 8 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 2 \ -1 \end{bmatrix} \].

Step 2: Find the Determinant of the Coefficient Matrix

The coefficient matrix is \( A = \begin{bmatrix} 10 & -6 \ -5 & 8 \end{bmatrix} \). To find its determinant, use the formula for a 2x2 matrix: \( \, \text{det}(A) = ad - bc \). Here, \( a = 10 \), \( b = -6 \), \( c = -5 \), \( d = 8 \). Calculate: \( \, \text{det}(A) = 10 \times 8 - (-6) \times (-5) = 80 - 30 = 50 \).

Step 3: Calculate the Determinant for x (Determinant D_x)

Create a new matrix \( A_x \) by replacing the first column of \( A \) with the constants from the right side of the equations: \[ A_x = \begin{bmatrix} 2 & -6 \ -1 & 8 \end{bmatrix} \] Find its determinant: \( \, \text{det}(A_x) = 2 \times 8 - (-6) \times (-1) = 16 + 6 = 22 \).

Step 4: Calculate the Determinant for y (Determinant D_y)

Create a new matrix \( A_y \) by replacing the second column of \( A \) with the constants from the right side of the equations: \[ A_y = \begin{bmatrix} 10 & 2 \ -5 & -1 \end{bmatrix} \] Find its determinant: \( \, \text{det}(A_y) = 10 \times (-1) - 2 \times (-5) = -10 + 10 = 0 \).

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

Cramer's Rule states: \( x = \frac{\text{det}(A_x)}{\text{det}(A)} \) and \( y = \frac{\text{det}(A_y)}{\text{det}(A)} \). We have: \( x = \frac{22}{50} = 0.44 \) and \( y = \frac{0}{50} = 0 \).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations involving the same set of variables. For example, in the given problem, we have:

• \(10x - 6y = 2\)
• \(-5x + 8y = -1\)

These equations are linear because each term is either a constant or a product of a constant and a single variable. Linear equations are used to determine variable values that satisfy all equations simultaneously.
To solve a system of linear equations, we often aim to find the values of \(x\) and \(y\) that satisfy both equations at once. This can be represented graphically as the point where the lines intersect each other.
When using algebraic methods like Cramer's Rule, systems are expressed in terms of matrices to utilize matrix arithmetic for solutions.

Matrix Determinants

Matrix determinants are scalars that provide important information about a matrix. They are central when solving systems of linear equations using methods such as Cramer's Rule. For a 2x2 matrix, the determinant is calculated as follows for a matrix \(A\):

• Suppose \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\).
• The determinant, expressed as \(\text{det}(A)\), is calculated by \(ad - bc\).

In the provided exercise, the coefficient matrix is:\[\begin{bmatrix} 10 & -6 \ -5 & 8 \end{bmatrix}\]Plugging into the determinant formula, we find \( \text{det}(A) = 10 \times 8 - (-6 \times -5) = 80 - 30 = 50 \).
The determinant indicates whether the system of equations has a unique solution, no solution, or infinitely many solutions. If the determinant is non-zero, as in our case, a unique solution exists.

Solving Algebraic Equations

Solving algebraic equations involves finding the values that satisfy the equation. With Cramer's Rule, this involves a systematic approach using determinants. Let’s break down how this rule applies:

• First, calculate the determinant of the coefficient matrix, \(\text{det}(A)\), to ensure a unique solution.
• Create matrices \(A_x\) and \(A_y\) by replacing their respective columns with the constant terms from the equations.
• Calculate the determinants \(\text{det}(A_x)\) and \(\text{det}(A_y)\).
• Apply Cramer's Rule: \( x = \frac{\text{det}(A_x)}{\text{det}(A)} \) and \( y = \frac{\text{det}(A_y)}{\text{det}(A)} \).

Using the problem's data, \( x \) was computed as \( \frac{22}{50} = 0.44 \) and \( y \) as \( \frac{0}{50} = 0 \).
This shows how linear algebra techniques can solve equations efficiently, giving precise solutions where other methods might be cumbersome.