Problem 52 - Precalculus

Question

Apply Cramer's rule to solve each system of equations, if possible. $$\begin{array}{r} x+y+z=6 \\ x-y-z=0 \\ -x+y+z=7 \end{array}$$

Step-by-Step Solution

Answer

The system cannot be solved using Cramer's rule as its determinant is zero, indicating no unique solution.

Step 1: Write the System in Matrix Form

To apply Cramer's rule, first express the system in matrix form, which is \( A\mathbf{x} = \mathbf{b} \).The coefficient matrix \( A \) is: \[A = \begin{bmatrix} 1 & 1 & 1 \ 1 & -1 & -1 \ -1 & 1 & 1 \end{bmatrix} \]The variables vector \( \mathbf{x} \) is:\[\mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \]And the constants vector \( \mathbf{b} \) is:\[\mathbf{b} = \begin{bmatrix} 6 \ 0 \ 7 \end{bmatrix} \]

Step 2: Calculate the Determinant of A

To use Cramer's rule, compute the determinant of the coefficient matrix \( A \). Let's denote this determinant as \( \text{det}(A) \).\[\text{det}(A) = \begin{vmatrix} 1 & 1 & 1 \ 1 & -1 & -1 \ -1 & 1 & 1 \end{vmatrix} \]Using cofactor expansion along the first row:\[\text{det}(A) = 1 \begin{vmatrix} -1 & -1 \ 1 & 1 \end{vmatrix} -1 \begin{vmatrix} 1 & -1 \ -1 & 1 \end{vmatrix} +1 \begin{vmatrix} 1 & -1 \ -1 & 1 \end{vmatrix} \]Calculate each smaller determinant:- \( \begin{vmatrix} -1 & -1 \ 1 & 1 \end{vmatrix} = (-1)(1) - (-1)(1) = 0 \)- \( \begin{vmatrix} 1 & -1 \ -1 & 1 \end{vmatrix} = (1)(1) - (-1)(-1) = 0 \)Substituting these into the expression gives:\[\text{det}(A) = 1(0) -1(0) + 1(0) = 0 \]

Step 3: Interpret the Determinant Result

Since \( \text{det}(A) = 0 \), the matrix \( A \) is singular and does not have an inverse. This implies that Cramer's rule cannot be applied to solve this system of equations because it does not have a unique solution. The system may have either no solutions or infinitely many solutions.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. These equations work together to define a relationship between their variables. In practical applications, solving such a system can help us find the values of those variables.
Consider the given system of equations in three variables:

• \(x + y + z = 6\)
• \(x - y - z = 0\)
• \(-x + y + z = 7\)

Each equation provides a constraint, and our task is to find if there is a point (or points) that satisfies all these constraints simultaneously.
When we attempt to solve this system using Cramer's Rule, a mathematical technique that can provide solutions if they exist, the result depends on the properties of the associated matrices, particularly their determinants.

Matrix Form

To effectively apply techniques like Cramer's Rule, it's crucial to express the system in matrix form. Matrices provide a compact representation of systems of equations, which aids in using various algebraic operations.
The general matrix form of a system \( A\mathbf{x} = \mathbf{b} \) consists of:

• **Matrix \( A \):** Contains the coefficients of the variables from each equation. For our system:\[A = \begin{bmatrix} 1 & 1 & 1 \1 & -1 & -1 \-1 & 1 & 1 \end{bmatrix}\]
• **Vector \( \mathbf{x} \):** Represents the column vector of variables: \[\mathbf{x} = \begin{bmatrix} x \y \z \end{bmatrix}\]
• **Vector \( \mathbf{b} \):** Captures the constants on the right-hand side of each equation: \[\mathbf{b} = \begin{bmatrix} 6 \0 \7 \end{bmatrix}\]

This matrix form helps simplify the problem and allows the use of linear algebra techniques, making the system easier to analyze and solve.

Determinant Calculation

The determinant is a special number calculated from a square matrix. It serves as an essential focus in solving systems using Cramer's Rule because it indicates if a unique solution exists.
For our matrix \( A \), the determinant, \( \text{det}(A) \), is calculated as follows:

• Apply cofactor expansion along any row or column. Often, the first row is used for simplicity.
• For matrix \( A \):\[\text{det}(A) = 1 \begin{vmatrix} -1 & -1 \1 & 1 \end{vmatrix} -1 \begin{vmatrix} 1 & -1 \-1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \-1 & 1 \end{vmatrix}\]

Calculate the smaller 2x2 determinants (called minors) for each part. Interestingly, in this case, each minor is zero, leading to \( \text{det}(A) = 0 \).
A zero determinant indicates that the matrix is singular and hence does not have an inverse. This means there are either no solutions or infinitely many solutions—implying the system is either dependent or inconsistent. Therefore, Cramer's Rule cannot be applied to find unique solutions, emphasizing the importance of the determinant in analyzing the system's solvability.