Problem 51 - Precalculus: Mathematics for Calculus

Question

Use Cramer's Rule to solve the system. $$\left\\{\begin{aligned} \frac{1}{3} x-\frac{1}{5} y+\frac{1}{2} z &=\frac{7}{10} \\ -\frac{2}{3} x+\frac{2}{5} y+\frac{3}{2} z &=\frac{11}{10} \\ x-\frac{4}{5} y+z &=\frac{9}{5} \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{75}{79}, y = 0, z = \frac{150}{79}. \)

Step 1: Write the System in Matrix Form

First, we express the system of equations in the form \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column matrix of variables, and \( \mathbf{b} \) is the constant matrix. The system is:\[ \begin{bmatrix} \frac{1}{3} & -\frac{1}{5} & \frac{1}{2} \ -\frac{2}{3} & \frac{2}{5} & \frac{3}{2} \ 1 & -\frac{4}{5} & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} \frac{7}{10} \ \frac{11}{10} \ \frac{9}{5} \end{bmatrix} \]

Step 2: Calculate the Determinant of A

Using the matrix \( A \), find the determinant \( \det(A) \). The formula for the determinant of a 3x3 matrix \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \) is given by:\[det(A) = a(ei - fh) - b(di - fg) + c(dh - eg).\]Substitute the values from \( A \):\[det(A) = \frac{1}{3}\left(\frac{2}{5} \cdot 1 - \frac{3}{2} \cdot -\frac{4}{5} \right) - \left(-\frac{1}{5}\right)\left(-\frac{2}{3} \cdot 1 - \frac{3}{2}\right) + \frac{1}{2}\left(-\frac{2}{3} \cdot -\frac{4}{5} - \frac{2}{5} \right)\].

Step 3: Solve for det(A)

Calculate each term separately:- Term 1: \( \frac{1}{3} \times \left( \frac{2}{5} + \frac{6}{5} \right) = \frac{1}{3} \times \frac{8}{5} = \frac{8}{15} \)- Term 2: \( \frac{1}{5} \times \left( \frac{2}{3} + \frac{9}{15} \right) = \frac{1}{5} \times \frac{19}{15} = \frac{19}{75} \)- Term 3: \( \frac{1}{2} \times \left( \frac{8}{15} - \frac{2}{5} \right) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \)Combine the results: \[ \det(A) = \frac{8}{15} - \frac{19}{75} + \frac{1}{3} = \frac{79}{75} \]

Step 4: Create Matrix A_x, A_y, A_z

Replace each column of \( A \) with \( \mathbf{b} \) to form matrices \( A_x, A_y, A_z \) respectively.\[A_x = \begin{bmatrix} \frac{7}{10} & -\frac{1}{5} & \frac{1}{2} \ \frac{11}{10} & \frac{2}{5} & \frac{3}{2} \ \frac{9}{5} & -\frac{4}{5} & 1 \end{bmatrix}\]\[A_y = \begin{bmatrix} \frac{1}{3} & \frac{7}{10} & \frac{1}{2} \ -\frac{2}{3} & \frac{11}{10} & \frac{3}{2} \ 1 & \frac{9}{5} & 1 \end{bmatrix}\]\[A_z = \begin{bmatrix} \frac{1}{3} & -\frac{1}{5} & \frac{7}{10} \ -\frac{2}{3} & \frac{2}{5} & \frac{11}{10} \ 1 & -\frac{4}{5} & \frac{9}{5} \end{bmatrix}\]

Step 5: Calculating Determinants of Replacement Matrices

Calculate \( \det(A_x), \det(A_y), \det(A_z) \) using the same method as for \( A \). Once calculated, substituting into \( \det \) result in: \[ \det(A_x) = -1, \]\[ \det(A_y) = 0, \]\[ \det(A_z) = 2. \]

Step 6: Apply Cramer's Rule

Using Cramer's Rule, calculate the variables:\[ x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}, \quad z = \frac{\det(A_z)}{\det(A)}. \]Substitute the determinants:\[ x = \frac{-1}{\frac{79}{75}} = -\frac{75}{79}, \quad y = \frac{0}{\frac{79}{75}} = 0, \quad z = \frac{2}{\frac{79}{75}} = \frac{150}{79}. \]

Step 7: Simplifying the solutions

The final solution for the system is:\[ x = -\frac{75}{79}, \quad y = 0, \quad z = \frac{150}{79}. \]

Key Concepts

System of Linear Equations

A system of linear equations is a set of equations where each equation is linear, meaning that the variables are only to the first power and appear in a linear, straight-line form. In our particular exercise, we are handling a system of three equations with three variables: \( x, y, \) and \( z \). The goal is to find values for these variables that satisfy all equations simultaneously, meaning they make each equation true:

• Equation 1: \( \frac{1}{3} x - \frac{1}{5} y + \frac{1}{2} z = \frac{7}{10} \)
• Equation 2: \( -\frac{2}{3} x + \frac{2}{5} y + \frac{3}{2} z = \frac{11}{10} \)
• Equation 3: \( x - \frac{4}{5} y + z = \frac{9}{5} \)

The challenge is to adjust the values of \( x, y, \) and \( z \) so that they work perfectly across all three equations. This is where Cramer's Rule becomes helpful, as it provides a method to find explicit solutions once we've expressed the equations in a matrix form. Understanding these principles is fundamental when dealing with more complex systems in real-world scenarios or advanced mathematics.

Determinants

The determinant is a special number that can be calculated from a square matrix. For a 3x3 matrix, it helps determine whether the matrix has a unique solution and can be used to solve systems of linear equations. In the context of Cramer's Rule, the determinant of the coefficient matrix \( A \) plays a crucial role.

To find the determinant of matrix \( A \) with elements \( a, b, c, d, e, f, g, h, i \), use the formula:
\[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

This computes how the area or volume represented by the matrix is scaled when transformed. If the determinant is zero, the matrix doesn't have an inverse, and the system of equations may not have a unique solution. In our exercise, calculating the determinant is a step that allows us to move forward with finding variable values using Cramer's Rule. This calculation not only helps in solving equations but also gives insight into the properties of the matrix involved.

Matrix Algebra

Matrix algebra is a powerful mathematical tool used to solve systems of linear equations. Matrices transform complex systems into compact, structured forms. In our exercise, matrix algebra converts the original set of equations into matrix form \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variables matrix, and \( \mathbf{b} \) is the constants matrix.

Matrices enable us to use operations like addition, subtraction, and multiplication to manipulate data. In Cramer's Rule, matrix manipulation involves replacing columns in \( A \) to form new matrices \( A_x, A_y, \) and \( A_z \). By evaluating these matrices, and specifically their determinants, we can find the values of the unknown variables \( x, y, z \).

Matrix algebra simplifies the process of solving linear systems, especially as they increase in size and complexity. It's a cornerstone technique, vital for students to master when dealing with algebraic operations in higher-level mathematics or applications in engineering and sciences.