Problem 66 - Intermediate Algebra

Question

Use Cramer's rule to solve each system of equations. If a system is inconsistent or if the equations are dependent, so indicate. $$ \left\\{\begin{array}{l} \frac{1}{2} x+y+z+\frac{3}{2}=0 \\ x+\frac{1}{2} y+z-\frac{1}{2}=0 \\ x+y+\frac{1}{2} z+\frac{1}{2}=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 0, y = -1, z = -\frac{1}{4} \).

Step 1: Convert into Standard Form

First, rewrite each equation so it is in standard form: \( ax + by + cz = d \). This gives us: \( \frac{1}{2}x + y + z = -\frac{3}{2} \), \( x + \frac{1}{2}y + z = \frac{1}{2} \), and \( x + y + \frac{1}{2}z = -\frac{1}{2} \).

Step 2: Write the Coefficient Matrix

The coefficient matrix \( A \) for the system is:\[ A = \begin{bmatrix} \frac{1}{2} & 1 & 1 \ 1 & \frac{1}{2} & 1 \ 1 & 1 & \frac{1}{2} \end{bmatrix} \]

Step 3: Write the Constant Matrix

The constants on the right side form the matrix \( B \):\[ B = \begin{bmatrix} -\frac{3}{2} \ \frac{1}{2} \ -\frac{1}{2} \end{bmatrix} \]

Step 4: Calculate the Determinant of A

To use Cramer's rule, compute the determinant of \( A \), denoted as \( \det(A) \): \[ \det(A) = \frac{1}{2}(\frac{1}{2}\cdot\frac{1}{2} - 1\cdot1) - 1(1\cdot\frac{1}{2} - 1\cdot1) + 1(1\cdot1 - \frac{1}{2}\cdot\frac{1}{2}) = -\frac{1}{4} + \frac{1}{2} + \frac{3}{4} = 1 \]

Step 5: Construct and Solve Determinants for Cramer's Rule

Create modified matrices \( A_x, A_y, A_z \) by replacing the respective columns of \( A \) with \( B \), and then calculate their determinants.- \( A_x = \begin{bmatrix} -\frac{3}{2} & 1 & 1 \ \frac{1}{2} & \frac{1}{2} & 1 \ -\frac{1}{2} & 1 & \frac{1}{2} \end{bmatrix} \),\( \det(A_x) = -\frac{3}{4} + \frac{1}{2} + \frac{1}{4} = 0 \).- \( A_y = \begin{bmatrix} \frac{1}{2} & -\frac{3}{2} & 1 \ 1 & \frac{1}{2} & 1 \ 1 & -\frac{1}{2} & \frac{1}{2} \end{bmatrix} \),\( \det(A_y) = -\frac{3}{4} + \frac{3}{4} -1 = -1 \).- \( A_z = \begin{bmatrix} \frac{1}{2} & 1 & -\frac{3}{2} \ 1 & \frac{1}{2} & \frac{1}{2} \ 1 & 1 & -\frac{1}{2} \end{bmatrix} \),\( \det(A_z) = -\frac{3}{4} - \frac{1}{2} + 1 = -\frac{1}{4} \).

Step 6: Apply Cramer's Rule

Using Cramer's rule, solve for \( x, y, z \) by dividing each determinant by \( \det(A) \):- \( x = \frac{\det(A_x)}{\det(A)} = \frac{0}{1} = 0 \)- \( y = \frac{\det(A_y)}{\det(A)} = \frac{-1}{1} = -1 \)- \( z = \frac{\det(A_z)}{\det(A)} = \frac{-\frac{1}{4}}{1} = -\frac{1}{4} \).

Key Concepts

Systems of Equations

A system of equations is a set of multiple equations with multiple variables. The goal is to find the values of these variables that satisfy all equations simultaneously. When dealing with systems, three situations can arise: there is a unique solution, there are infinitely many solutions, or there is no solution at all.

In the given exercise, we have 3 equations with 3 variables: \( x, y, \) and \( z \). Each equation represents a plane in three-dimensional space. Solving this system aims to find the point of intersection of these planes. Here’s a reminder of how to write each equation:

• Standard form: \( ax + by + cz = d \).
• The equations need to be manipulated into this form before using techniques like Cramer’s Rule.

The standard form makes it easier to apply matrix operations, which is critical for solving the system using algebraic methods like Cramer's Rule.

Determinant Calculation

The determinant is a special number that can be calculated from a square matrix. It provides valuable information about the matrix, such as whether a unique solution exists for the system of equations. A non-zero determinant indicates a unique solution.

To calculate the determinant of a 3x3 matrix, use the formula:
\[ \det \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

In the problem's solution, the determinant of the original coefficient matrix \( A \) was computed step by step. Each element's contribution to the determinant was calculated by considering the minor matrix formed by removing the row and column of the current element.

• Calculate for any element as: element × determinant(minor).
• Alternate signs between positive and negative.

This calculation verifies whether Cramer’s Rule is applicable by confirming that \( \det(A) eq 0 \). A non-zero value confirms a unique solution exists for the given system.

Matrix Algebra

Matrix algebra is a branch of mathematics that deals with matrices, arrays of numbers or variables arranged in rows and columns. When working with systems of equations, matrix algebra allows us to represent the system and efficiently solve it using methods such as Cramer's Rule.

In the exercise, the system is represented using matrices:

• Coefficient matrix \( A \): encapsulates coefficients of variables from the system equations.
• Constant matrix \( B \): contains constants from the right side of each equation.

The coefficient matrix \( A \) helps in organizing and simplifying the problem. By applying methods from linear algebra, we use matrix operations to find solutions. Cramer's Rule, specifically, requires calculating determinants of these matrices to find the variable solutions:

• Replace columns in the coefficient matrix with the constant matrix to form new matrices like \( A_x, A_y, \) and \( A_z \).
• Perform determinant calculations on these modified matrices to solve for each variable using the formula \( x = \frac{\det(A_x)}{\det(A)} \).

This streamlines the process of solving complex systems and showcases the power of matrix algebra in handling and simplifying algebraic problems.