Problem 32 - College Algebra with Modeling and Visualization

Question

Use Cramer's rule to solve the system of linear equations. $$ \begin{aligned} -2.7 x+1.5 y &=-1.53 \\ 1.8 x-5.5 y &=-1.68 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( x \approx 0.9 \), \( y \approx 0.6 \).

Step 1: Write the System in Matrix Form

Express the system of equations as a matrix equation of the form \( AX = B \). The matrix \( A \), the vector \( X \), and the vector \( B \) are as follows:\[ A = \begin{bmatrix} -2.7 & 1.5 \ 1.8 & -5.5 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} -1.53 \ -1.68 \end{bmatrix} \]

Step 2: Calculate the Determinant of Matrix A

The determinant of matrix \( A \), denoted as \( \det(A) \), is calculated as:\[ \det(A) = (-2.7)(-5.5) - (1.8)(1.5) \]Simplify it to find the determinant value.

Step 3: Compute \( \det(A) \) Value

Calculate \( \det(A) \):\[ \det(A) = 14.85 - 2.7 = 12.15 \]

Step 4: Formulate Matrix A_x and Calculate its Determinant

Replace the first column of matrix \( A \) with vector \( B \) to get \( A_x \):\[ A_x = \begin{bmatrix} -1.53 & 1.5 \ -1.68 & -5.5 \end{bmatrix} \]Calculate \( \det(A_x) \):\[ \det(A_x) = (-1.53)(-5.5) - (-1.68)(1.5) \]

Step 5: Compute \( \det(A_x) \) Value

Calculate \( \det(A_x) \):\[ \det(A_x) = 8.415 + 2.52 = 10.935 \]

Step 6: Formulate Matrix A_y and Calculate its Determinant

Replace the second column of matrix \( A \) with vector \( B \) to get \( A_y \):\[ A_y = \begin{bmatrix} -2.7 & -1.53 \ 1.8 & -1.68 \end{bmatrix} \]Calculate \( \det(A_y) \):\[ \det(A_y) = (-2.7)(-1.68) - (1.8)(-1.53) \]

Step 7: Compute \( \det(A_y) \) Value

Calculate \( \det(A_y) \):\[ \det(A_y) = 4.536 + 2.754 = 7.29 \]

Step 8: Solve for x and y Using Cramer’s Rule

According to Cramer's Rule, solve for \( x \) and \( y \) as follows:\[ x = \frac{\det(A_x)}{\det(A)} = \frac{10.935}{12.15} \approx 0.9 \]\[ y = \frac{\det(A_y)}{\det(A)} = \frac{7.29}{12.15} \approx 0.6 \]

Step 9: Conclude the Solution

The solution to the system of equations is \( x \approx 0.9 \) and \( y \approx 0.6 \).

Key Concepts

System of Linear Equations

In algebra, a **system of linear equations** refers to a set of equations that involve two or more variables where each equation represents a straight line. In our exercise, we are given two equations with two variables, x and y:\[\begin{aligned} -2.7x + 1.5y &= -1.53 \1.8x - 5.5y &= -1.68\end{aligned}\] These equations intersect because they both must satisfy the same condition for the variables x and y.
To solve such a system graphically, you’d typically plot both equations on the coordinate axis and look for the point where they intersect. However, modern algebra allows us to solve these analytically using various methods, including Cramer's Rule.
Understanding systems of equations is crucial since they are commonly used in real-world applications, such as economics for supply and demand analysis or physics in understanding forces acting upon an object.

Determinants

The concept of **determinants** plays a fundamental role in matrix algebra. A determinant provides a scalar value derived from a square matrix and is used primarily to determine if a matrix is invertible.
In our exercise, we calculate the determinant of the matrix \( A \) given by:\[A = \begin{bmatrix} -2.7 & 1.5 \1.8 & -5.5 \end{bmatrix} \]
The determinant, denoted as \( \det(A) \), is calculated for a 2x2 matrix as:\[\det(A) = ad - bc \]
where \( a = -2.7 \), \( b = 1.5 \), \( c = 1.8 \), and \( d = -5.5 \). This value is integral to solving the system using Cramer’s Rule. Matrix \( A \) has a determinant \( \det(A) = 12.15 \) in our solution. If the determinant had been zero, Cramer's Rule wouldn't apply as it requires a non-zero determinant for a unique solution. Determinants extend beyond solving systems, finding applications in calculus and differential equations.

Matrix Algebra

**Matrix algebra** is a collection of mathematical techniques used for dealing with matrices, which are rectangular arrays of numbers. In our problem, the system of equations was expressed in matrix form to facilitate the use of Cramer's Rule.
A matrix equation can be expressed as \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the column matrix containing the unknown variables, and \( B \) is the constant matrix.
In step-by-step solutions, we rearrange and manipulate these matrices to isolate the variables.

• The coefficient matrix \( A \) is manipulated to form new matrices \( A_x \) and \( A_y \), crucial for calculating the unique solutions for x and y in Cramer’s Rule.
• Matrix operations such as multiplication and finding inverses are key components of matrix algebra.

This particular problem illustrates how structured algebraic approaches like matrix algebra can efficiently solve complex variable systems by simplifying them into linear, standardized processes.