Problem 65 - 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. $$ \left\\{\begin{array}{l} x+y=1 \\ \frac{1}{2} y+z=\frac{5}{2} \\ x-z=-3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 0 \), \( y = -\frac{4}{3} \), \( z = -\frac{2}{3} \).

Step 1: Write the System in Matrix Form

To use Cramer's rule, we first write the system of equations in matrix form. The system is already in appropriate form, allowing us to extract coefficient, variable, and constant matrices. The coefficient matrix \( A \), the variable matrix \( X \), and the constant matrix \( B \) are given by: \[ A = \begin{bmatrix} 1 & 1 & 0 \ 0 & \frac{1}{2} & 1 \ 1 & 0 & -1 \end{bmatrix}, \, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \, B = \begin{bmatrix} 1 \ \frac{5}{2} \ -3 \end{bmatrix} \] The system can be expressed as \( A \cdot X = B \).

Step 2: Calculate the Determinant of Coefficient Matrix (\( |A| \))

Compute the determinant of the coefficient matrix \( A \). \( |A| = \begin{vmatrix} 1 & 1 & 0 \ 0 & \frac{1}{2} & 1 \ 1 & 0 & -1 \end{vmatrix} \) Using cofactor expansion along the first row, we calculate: \[ |A| = 1 \left( \begin{vmatrix} \frac{1}{2} & 1 \ 0 & -1 \end{vmatrix} \right) - 1 \left( \begin{vmatrix} 0 & 1 \ 1 & -1 \end{vmatrix} \right) \]Evaluating the minors:\[ \begin{vmatrix} \frac{1}{2} & 1 \ 0 & -1 \end{vmatrix} = \left(\frac{1}{2} \times -1 - 0\right) = -\frac{1}{2}, \quad \begin{vmatrix} 0 & 1 \ 1 & -1 \end{vmatrix} = \left(0 + 1\right) = 1 \]Thus, the determinant is: \[ |A| = 1(-\frac{1}{2}) - 1(1) = -\frac{1}{2} - 1 = -\frac{3}{2} \]Since \( |A| eq 0 \), the system has a unique solution.

Step 3: Calculate Determinants for Variables

Calculate the determinant for each variable by replacing the respective column of \( A \) with \( B \), and find their values: For \( x \), replace the first column of \( A \) with \( B \):\[ A_x = \begin{bmatrix} 1 & 1 & 0 \ \frac{5}{2} & \frac{1}{2} & 1 \ -3 & 0 & -1 \end{bmatrix} \] \( |A_x| = \begin{vmatrix} 1 & 1 & 0 \ \frac{5}{2} & \frac{1}{2} & 1 \ -3 & 0 & -1 \end{vmatrix} \) Using first row cofactor expansion:\[ |A_x| = 1 \left( \begin{vmatrix} \frac{1}{2} & 1 \ 0 & -1 \end{vmatrix} \right) - 1 \left( \begin{vmatrix} \frac{5}{2} & 1 \ -3 & -1 \end{vmatrix} \right) \]Minors are evaluated to:\[ \begin{vmatrix} \frac{1}{2} & 1 \ 0 & -1 \end{vmatrix} = -\frac{1}{2}, \quad \begin{vmatrix} \frac{5}{2} & 1 \ -3 & -1 \end{vmatrix} = -\frac{5}{2} + 3 = -\frac{1}{2} \]Thus, the determinant is:\[ |A_x| = 1(-\frac{1}{2}) - 1(-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0 \] For \( y \), replace the second column of \( A \) with \( B \):\[ A_y = \begin{bmatrix} 1 & 1 & 0 \ 0 & \frac{5}{2} & 1 \ 1 & -3 & -1 \end{bmatrix} \] \( |A_y| = \begin{vmatrix} 1 & 1 & 0 \ 0 & \frac{5}{2} & 1 \ 1 & -3 & -1 \end{vmatrix} \) Using first row cofactor expansion:\[ |A_y| = 1 \left( \begin{vmatrix} \frac{5}{2} & 1 \ -3 & -1 \end{vmatrix} \right) - 1 \left( \begin{vmatrix} 0 & 1 \ 1 & -1 \end{vmatrix} \right) \]Minors are evaluated to:\[ \begin{vmatrix} \frac{5}{2} & 1 \ -3 & -1 \end{vmatrix} = 1, \quad \begin{vmatrix} 0 & 1 \ 1 & -1 \end{vmatrix} = -1 \]Thus, the determinant is:\[ |A_y| = 1(1) - 1(-1) = 1 + 1 = 2 \] For \( z \), replace the third column of \( A \) with \( B \):\[ A_z = \begin{bmatrix} 1 & 1 & 1 \ 0 & \frac{1}{2} & \frac{5}{2} \ 1 & 0 & -3 \end{bmatrix} \] \( |A_z| = \begin{vmatrix} 1 & 1 & 1 \ 0 & \frac{1}{2} & \frac{5}{2} \ 1 & 0 & -3 \end{vmatrix} \) Using first row cofactor expansion:\[ |A_z| = 1 \left( \begin{vmatrix} \frac{1}{2} & \frac{5}{2} \ 0 & -3 \end{vmatrix} \right) - 1 \left( \begin{vmatrix} 0 & \frac{5}{2} \ 1 & -3 \end{vmatrix} \right) + 1 \left( \begin{vmatrix} 0 & \frac{1}{2} \ 1 & 0 \end{vmatrix} \right) \]Minors are evaluated to:\[ \begin{vmatrix} \frac{1}{2} & \frac{5}{2} \ 0 & -3 \end{vmatrix} = -\frac{3}{2}, \quad \begin{vmatrix} 0 & \frac{5}{2} \ 1 & -3 \end{vmatrix} = -\frac{5}{2}, \quad \begin{vmatrix} 0 & \frac{1}{2} \ 1 & 0 \end{vmatrix} = -\frac{1}{2} \]Thus, the determinant is:\[ |A_z| = 1(-\frac{3}{2}) - 1(-\frac{5}{2}) + 1(-\frac{1}{2}) = -\frac{3}{2} + \frac{5}{2} - \frac{1}{2} = 1 \]

Step 4: Apply Cramer's Rule

Using Cramer's Rule, each variable is found by dividing its determinant by \( |A| \). Therefore: \[ x = \frac{|A_x|}{|A|} = \frac{0}{-\frac{3}{2}} = 0 \]\[ y = \frac{|A_y|}{|A|} = \frac{2}{-\frac{3}{2}} = -\frac{4}{3} \]\[ z = \frac{|A_z|}{|A|} = \frac{1}{-\frac{3}{2}} = -\frac{2}{3} \]

Key Concepts

Systems of Equations

In mathematics, a system of equations is a collection of two or more equations with a common set of unknowns. The objective is to find values for these unknowns that satisfy all the equations simultaneously. In our exercise, the system of equations is:

• \( x + y = 1 \)
• \( \frac{1}{2} y + z = \frac{5}{2} \)
• \( x - z = -3 \)

To solve such a system, we can use various methods, including substitution, elimination, or matrix methods like Cramer's rule. Cramer's rule is particularly powerful for linear systems that can be expressed in matrix form, having non-zero determinants. This method is beneficial when dealing with systems that need precise calculation of multiple variables.

Determinants

A determinant is a special number calculated from a square matrix. It provides crucial information about the matrix, such as whether it has an inverse and whether the system of equations it represents has a unique solution, infinite solutions, or no solution.To use Cramer's rule, you must first compute the determinant of the coefficient matrix \( A \). In our exercise, the determinant \( |A| \) was calculated to be \(-\frac{3}{2}\). This value is not zero, indicating that the system of equations has a unique solution. If the determinant were zero, it would suggest that the equations are either dependent or inconsistent.Calculating the determinant involves cofactor expansion, which requires selecting a row or column, removing it, and calculating the minor determinants of the remaining elements, adjusting with signs.

Matrix Algebra

Matrix algebra is the study of matrices and their operations. It plays a vital role in solving systems of linear equations, transforming complex problems into simpler matrix operations.In matrix form, our system of equations is written as \( A \cdot X = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constant matrix. The beauty of matrix algebra lies in its ability to provide systematic solutions through operations such as matrix multiplication, finding inverses, and determinants.Using matrices allows for a compact and efficient way of handling multiple equations. Matrix methods like Cramer's rule utilize these algebraic properties to find unknowns efficiently by dealing directly with matrices and their determinants.

Linear Independence

Linear independence is a concept in linear algebra that indicates that a set of vectors or equations are non-redundant. In the context of systems of equations, it implies that no equation in the system can be derived from the others by manipulation.For our exercise, checking linear independence ensures that the system of equations is neither redundant nor inconsistent. Since the determinant \( |A| eq 0 \), we have a linearly independent system with a unique solution.Linear independence can often be validated by analyzing the determinant of the coefficient matrix. A non-zero determinant signifies that the system is full-rank, indicating that all equations present are necessary and contribute uniquely to the solution.