Problem 18 - Elementary and Intermediate Algebra

Question

Solve using Cramer's rule. $$ \begin{aligned} -2 x+4 y &=3 \\ 3 x-7 y &=1 \end{aligned} $$

Step-by-Step Solution

Answer

x = -12.5, y = -5.5

Step 1 - Write the system of equations in matrix form

Write the system of equations using a matrix representation.ewline\[\begin{aligned} \begin{pmatrix} -2 & 4 \ 3 & -7 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 3 \ 1 \end{pmatrix} \end{aligned}\]

Step 2 - Calculate the determinant of the coefficient matrix

The coefficient matrix is \[\begin{pmatrix} -2 & 4 \ 3 & -7 \end{pmatrix}\]Calculate its determinant by using \[\text{det}(A) = (-2 \cdot -7) - (4 \cdot 3) = 14 - 12 = 2\]So, \[\text{det}(A) = 2\]

Step 3 - Form the matrices for Cramer's Rule

To find the values of x and y, form the matrices by replacing the respective column of the coefficient matrix with the constant vector.ewlineFor x: \[A_x = \begin{pmatrix} 3 & 4 \ 1 & -7 \end{pmatrix}\]For y: \[A_y = \begin{pmatrix} -2 & 3 \ 3 & 1 \end{pmatrix}\]

Step 4 - Calculate the determinants of the new matrices

Calculate the determinants of the new matrices A_x and A_y.ewline\text{det}(A_x): \[\text{det}(A_x) = ( 3 \cdot -7 ) - ( 4 \cdot 1 ) = -21 - 4 = -25\] ewline\text{det}(A_y): \[\text{det}(A_y) = (-2 \cdot 1) - (3 \cdot 3) = -2 - 9 = -11\]

Step 5 - Apply Cramer's Rule to find the values of x and y

Using Cramer's Rule, solve for x and y.ewline\text{x}:\[x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-25}{2} = -12.5\]ewline\text{y}:\[y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-11}{2} = -5.5\]

Key Concepts

determinant

The determinant is a special number that can be calculated from a square matrix. It gives us important information about the matrix. For a 2x2 matrix, the determinant is calculated using the formula: \( \text{det}(A) = ad - bc \) where \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \).

Determinants are useful in solving systems of equations using Cramer's Rule. They tell us whether a system has a unique solution. If the determinant is zero, the system doesn't have a single unique solution. In the context of our exercise, we find the determinant of the coefficient matrix to determine potential solutions.

matrix form

Representing a system of equations in matrix form helps streamline the solution process. A system of equations can be written in a compact form using matrices. For example, our system: \(\-2x + 4y = 3 \) and \( 3x - 7y = 1 \), can be rewritten as: \begin{pmatrix} -2 & 4 \ 3 & -7 \end{pmatrix}\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 3 \ 1 \end{pmatrix}.
This matrix equation is in the form \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the column vector of variables, and \( B \) is the constant vector. This form makes it easier to apply mathematical techniques like Cramer's Rule.

system of equations

A system of equations is a set of equations with multiple variables. The goal is to find values for the variables that satisfy all equations simultaneously. For our exercise, the system of linear equations is: \(\-2x + 4y = 3 \) and \( 3x - 7y = 1 \).
These equations can be solved using various methods including substitution, elimination, and matrix methods like Cramer's Rule. Using a matrix approach converts the problem into a structured format that makes it easier to handle and solve systematically.

linear algebra

Linear algebra is a branch of mathematics dealing with vectors, matrices, and systems of linear equations. It includes concepts like linear transformations, vector spaces, and eigenvalues.
In the context of our exercise, linear algebra provides the tools like matrices and determinants to solve systems of equations efficiently. Cramer's Rule, which uses determinants, is a direct application of linear algebra. Understanding the fundamentals of linear algebra is crucial for solving more complex mathematical problems involving multiple variables.