Problem 11
Question
Use Cramer’s Rule to solve the system of linear equations. $$ \left\\{\begin{aligned} x+y &=8000 \\ 0.03 x+0.05 y &=250 \end{aligned}\right. $$
Step-by-Step Solution
Verified Answer
The solution is \( x = 7500 \) and \( y = 500 \).
1Step 1: Write the System in Matrix Form
Convert the given system of equations into a matrix form. The system is:\[ \begin{aligned} x + y &= 8000 \ 0.03x + 0.05y &= 250 \end{aligned} \] This can be written as \( AX = B \), where \( A = \begin{bmatrix} 1 & 1 \ 0.03 & 0.05 \end{bmatrix} \), \( X = \begin{bmatrix} x \ y \end{bmatrix} \), and \( B = \begin{bmatrix} 8000 \ 250 \end{bmatrix} \).
2Step 2: Find Determinant of A
Compute the determinant of matrix \( A \). The formula for the determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \( ad - bc \). For matrix \( A \), \( |A| = (1)(0.05) - (0.03)(1) = 0.05 - 0.03 = 0.02 \).
3Step 3: Formulate Determinant for x
Create matrix \( A_x \) by replacing the first column of \( A \) with \( B \). Thus \( A_x = \begin{bmatrix} 8000 & 1 \ 250 & 0.05 \end{bmatrix} \). Compute the determinant, \( |A_x| = (8000)(0.05) - (1)(250) = 400 - 250 = 150 \).
4Step 4: Formulate Determinant for y
Create matrix \( A_y \) by replacing the second column of \( A \) with \( B \). Thus \( A_y = \begin{bmatrix} 1 & 8000 \ 0.03 & 250 \end{bmatrix} \). Compute the determinant, \( |A_y| = (1)(250) - (0.03)(8000) = 250 - 240 = 10 \).
5Step 5: Solve for x and y Using Cramer's Rule
Apply Cramer's Rule, which states: \( x = \frac{|A_x|}{|A|} \) and \( y = \frac{|A_y|}{|A|} \). Therefore, \( x = \frac{150}{0.02} = 7500 \) and \( y = \frac{10}{0.02} = 500 \).
Key Concepts
Linear EquationsMatrix DeterminantSystem of EquationsMatrix Formulation
Linear Equations
Linear Equations are rules that involve constants, variables, and arithmetic operations, like addition and multiplication. These equations often show how different variables are connected.
In the context of systems of linear equations, you have multiple equations that must be solved together to find values of variables.
For example, if we have two linear equations with two variables, such as:
In the context of systems of linear equations, you have multiple equations that must be solved together to find values of variables.
For example, if we have two linear equations with two variables, such as:
- \(x + y = 8000\)
- \(0.03x + 0.05y = 250\)
Matrix Determinant
A Matrix Determinant is a special value that can be calculated from a square matrix. It provides insightful details about the matrix, like if it's invertible or the orientation of vectors when transformed by the matrix.
The determinant for a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated as:
The determinant for a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated as:
- \(ad - bc\)
- \( \begin{bmatrix} 1 & 1 \ 0.03 & 0.05 \end{bmatrix} \)
System of Equations
A System of Equations is a collection of two or more equations with the same set of unknowns. The goal is to find the values of the variables that satisfy all equations in the system.
In our example, the system consists of the following equations:
In our example, the system consists of the following equations:
- \(x + y = 8000\)
- \(0.03x + 0.05y = 250\)
Matrix Formulation
Matrix Formulation is a method of organizing systems of linear equations into a compact form using matrices. This approach allows the use of various matrix operations, such as finding determinants and inverses, to solve systems of equations efficiently.
In our case, the system of equations involving \(x\) and \(y\) can be represented in the form \(AX = B\) where:
In our case, the system of equations involving \(x\) and \(y\) can be represented in the form \(AX = B\) where:
- \(A\) is the coefficient matrix: \( \begin{bmatrix} 1 & 1 \ 0.03 & 0.05 \end{bmatrix} \)
- \(X\) is the variable matrix: \( \begin{bmatrix} x \ y \end{bmatrix} \)
- \(B\) is the constant matrix: \( \begin{bmatrix} 8000 \ 250 \end{bmatrix} \)
Other exercises in this chapter
Problem 11
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