Problem 27 - Intermediate Algebra

Question

For Problems \(11-30\), use Cramer's rule to find the solution set of each system. (Objective 2) $$ \left(\begin{array}{rr} -x-y+5 z= & 4 \\ x+y-7 z= & -6 \\ 2 x+3 y+4 z= & 13 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution set is \((0, 0, 1)\).

Step 1: Write the System in Matrix Form

First, write the given system of equations in the matrix form. It becomes: \[ A\mathbf{x} = \mathbf{b} \], where \( A = \begin{bmatrix} -1 & -1 & 5 \ 1 & 1 & -7 \ 2 & 3 & 4 \end{bmatrix} \), \( \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} 4 \ -6 \ 13 \end{bmatrix} \).

Step 2: Calculate the Determinant of A

Find the determinant of matrix \( A \). Use the formula for a 3x3 determinant: \[ \text{det}(A) = -1 (1 \cdot 4 - 3 \cdot (-7)) - (-1) (1 \cdot 4 - 2 \cdot (-7)) + 5 (1 \cdot 3 - 2 \cdot 1). \] Calculating this determinant yields \( \text{det}(A) = 119 \).

Step 3: Substitute to Find Determinant D_x

Substitute the first column of \( A \) with \( \mathbf{b} \) to get matrix \( A_x \): \[ A_x = \begin{bmatrix} 4 & -1 & 5 \ -6 & 1 & -7 \ 13 & 3 & 4 \end{bmatrix} \]. Calculate \( \text{det}(A_x) \), using the same determinant formula as before, which results in \( \text{det}(A_x) = 0 \).

Step 4: Substitute to Find Determinant D_y

Similarly, substitute the second column of \( A \) with \( \mathbf{b} \) to produce \( A_y \): \[ A_y = \begin{bmatrix} -1 & 4 & 5 \ 1 & -6 & -7 \ 2 & 13 & 4 \end{bmatrix} \]. Calculate \( \text{det}(A_y) \), which results in \( \text{det}(A_y) = 0 \).

Step 5: Substitute to Find Determinant D_z

Substitute the third column of \( A \) with \( \mathbf{b} \) to form \( A_z \): \[ A_z = \begin{bmatrix} -1 & -1 & 4 \ 1 & 1 & -6 \ 2 & 3 & 13 \end{bmatrix} \]. Calculate \( \text{det}(A_z) \), which results in \( \text{det}(A_z) = 119 \).

Step 6: Apply Cramer's Rule

Using Cramer's Rule: \( x = \frac{\text{det}(A_x)}{\text{det}(A)} \), \( y = \frac{\text{det}(A_y)}{\text{det}(A)} \), and \( z = \frac{\text{det}(A_z)}{\text{det}(A)} \). This gives \( x = 0 \), \( y = 0 \), \( z = 1 \).

Step 7: Present the Solution Set

The solution set for the system of equations is \( (x, y, z) = (0, 0, 1) \).

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. In this case, we have the equations:

• -x - y + 5z = 4
• x + y - 7z = -6
• 2x + 3y + 4z = 13

The goal is to find values for the variables that satisfy all equations simultaneously. Such problems often arise in various fields, like engineering and science, where it is essential to understand multiple interacting phenomena represented by these equations.
To solve these equations using Cramer's Rule, you first represent them in a matrix format, simplifying operations and calculations.

Matrix Determinant

The determinant of a matrix is a special number that can be calculated from its elements. It provides valuable information about the matrix, such as whether it has an inverse. For a 3x3 matrix like matrix \( A \): \[A = \begin{bmatrix} -1 & -1 & 5 \ 1 & 1 & -7 \ 2 & 3 & 4 \end{bmatrix}\]The determinant is computed using specific algebraic formulas. In this case, \( \text{det}(A) = 119 \).
The determinant is crucial in Cramer's Rule because it helps determine whether a unique solution exists for the system of equations. If the determinant is zero, the system might be "singular" or have no unique solution.

Linear Algebra

Linear algebra focuses on vector spaces and linear mappings between those spaces. It is fundamental to solving systems of equations. Through matrices and their operations, linear algebra provides a structured way to handle and solve equations.
By arranging our system as matrix \( A \mathbf{x} = \mathbf{b} \), we utilize powerful linear algebra techniques. This representation allows us to apply Cramer's Rule, which is a direct method for finding the solution of a linear system if the matrix determinant is non-zero. This is an often-used tool for proving the existence and finding the actual values of the solution set.

Solution Set

The solution set of a system of equations is the collection of all possible values for the variables that satisfy every equation in the system. When applying Cramer's Rule, once we compute the necessary determinants, each variable is given by:

• \( x = \frac{\text{det}(A_x)}{\text{det}(A)} \)
• \( y = \frac{\text{det}(A_y)}{\text{det}(A)} \)
• \( z = \frac{\text{det}(A_z)}{\text{det}(A)} \)

In this exercise, the solution set for the variables was determined to be \((0, 0, 1)\). This means that when \( x = 0 \), \( y = 0 \), and \( z = 1 \), all three equations in the system hold true, representing the unique solution to this particular system.