Problem 28 - Intermediate Algebra

Question

For Problems \(11-30\), use Cramer's rule to find the solution set of each system. (Objective 2) $$ \left(\begin{array}{rr} x+7 y-z= & -1 \\ -x-9 y+z= & 3 \\ 3 x+4 y-6 z= & 5 \end{array}\right) $$

Step-by-Step Solution

Answer

Solution: \((x, y, z) = (-27, 8, 58)\).

Step 1: Identify Matrices

First, identify the matrices involved in the system of equations. The system is:\[\begin{align*}x + 7y - z &= -1 \-x - 9y + z &= 3 \3x + 4y - 6z &= 5\end{align*}\]From this, the coefficient matrix \(A\) is:\[A = \begin{pmatrix} 1 & 7 & -1 \ -1 & -9 & 1 \ 3 & 4 & -6 \end{pmatrix}\]The variables matrix \(\mathbf{x}\) is:\[\mathbf{x} = \begin{pmatrix} x \ y \ z \end{pmatrix}\]And the constants matrix \(\mathbf{b}\) is:\[\mathbf{b} = \begin{pmatrix} -1 \ 3 \ 5 \end{pmatrix}\]

Step 2: Calculate the Determinant of Matrix A

To use Cramer's rule, we need the determinant of the coefficient matrix \(A\). Calculate the determinant \(\det(A)\) as follows:\[\det(A) = 1( (-9)(-6) - 1 \cdot 4) - 7( (-1)(-6) - 1\cdot 3) + (-1)( (-1)(4) - (-9)(3))\]Simplifying further:\[\det(A) = 1(54 - 4) - 7(6 - 3) - (4 + 27)\]\[\det(A) = 1(50) - 7(3) - 31\]\[\det(A) = 50 - 21 - 31 = -2\]The determinant \(\det(A) = -2\).

Step 3: Calculate Determinants for Matrices with Substituted Columns

Using Cramer's rule, calculate the determinants of matrices obtained by replacing columns of \(A\) with \(\mathbf{b}\).1. \(\det(A_x)\): Replace the first column of \(A\) with \(\mathbf{b}\):\[A_x = \begin{pmatrix} -1 & 7 & -1 \ 3 & -9 & 1 \ 5 & 4 & -6 \end{pmatrix}\]Calculate \(\det(A_x)\):\[\det(A_x) = -1((-9)(-6) - 1 \cdot 4) - 7(3 \cdot -6 - 1 \cdot 5) + (-1)(3 \cdot 4 - (-9) \cdot 5)\]\[\det(A_x) = -1(54 - 4) - 7(-18 - 5) - (12 + 45)\]\[\det(A_x) = -50 + 161 - 57 = 54\]2. \(\det(A_y)\): Replace the second column of \(A\) with \(\mathbf{b}\):\[A_y = \begin{pmatrix} 1 & -1 & -1 \ -1 & 3 & 1 \ 3 & 5 & -6 \end{pmatrix}\]Calculate \(\det(A_y)\):\[\det(A_y) = 1(3 \cdot -6 - 1 \cdot 5) + 1(-1 \cdot -6 - 1 \cdot 3) - 1(-1 \cdot 5 - 3 \cdot 3)\]\[\det(A_y) = 1(-18 - 5) + 1(6 - 3) - (5 - 9)\]\[\det(A_y) = -23 + 3 + 4 = -16\]3. \(\det(A_z)\): Replace the third column of \(A\) with \(\mathbf{b}\):\[A_z = \begin{pmatrix} 1 & 7 & -1 \ -1 & -9 & 3 \ 3 & 4 & 5 \end{pmatrix}\]Calculate \(\det(A_z)\):\[\det(A_z) = 1((-9)(5) - 3 \cdot 4) - 7(-1 \cdot 5 - 3 \cdot 3) + (-1)(-1 \cdot 4 - (-9) \cdot 3)\]\[\det(A_z) = 1(-45 - 12) + 7(5 - 9) - (4 + 27)\]\[\det(A_z) = -57 - 28 - 31 = -116\]

Step 4: Apply Cramer's Rule

With the determinants calculated, apply Cramer's rule to find the solutions for \(x\), \(y\), and \(z\):1. For \(x\):\[x = \frac{\det(A_x)}{\det(A)} = \frac{54}{-2} = -27\]2. For \(y\):\[y = \frac{\det(A_y)}{\det(A)} = \frac{-16}{-2} = 8\]3. For \(z\):\[z = \frac{\det(A_z)}{\det(A)} = \frac{-116}{-2} = 58\]Thus, the solution set is \((x, y, z) = (-27, 8, 58)\).

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. In our given exercise, we're dealing with a system that involves three equations and three variables: \(x\), \(y\), and \(z\). For instance:

• \(x + 7y - z = -1\)
• \(-x - 9y + z = 3\)
• \(3x + 4y - 6z = 5\)

The main goal is to find values for \(x, y,\) and \(z\) that will satisfy all three equations simultaneously. There are various methods to solve these systems, including substitution, elimination, and matrix methods like Cramer's Rule. Cramer's Rule uses determinants to find the solution, focusing on systems that can be expressed in square matrix form. This method is particularly helpful for understanding the structure of linear algebra.

Determinants

Determinants are numerical values derived from square matrices and are crucial in solutions involving linear algebra, such as using Cramer's Rule to solve a system of equations. The determinant provides insight into the properties of a matrix:

• If the determinant is zero, the matrix is singular, indicating that the system may not have a unique solution or could be dependent.
• If the determinant is non-zero, the matrix is invertible, suggesting the system has a unique solution.

Calculating the determinant of a 3x3 matrix like \(A = \begin{pmatrix} 1 & 7 & -1 \ -1 & -9 & 1 \ 3 & 4 & -6 \end{pmatrix}\) involves expanding along rows or columns using a formula that includes multiplying and summing specific elements and their minors. For matrix \(A\), the determinant was calculated as \(-2\), a non-zero value, ensuring that a unique solution to the system can be found using Cramer's Rule.

Linear Algebra

Linear algebra is a branch of mathematics focused on vector spaces and linear mappings between these spaces. It involves vectors, matrices, systems of linear equations, and transformations. In the context of our exercise, linear algebra encapsulates the structure and solution process for the system of equations using Cramer's Rule.A matrix is a core concept in linear algebra, representing the coefficients in a compact, manageable way. For the system of equations:\[\begin{align*}x + 7y - z =& -1\-x - 9y + z =& 3\3x + 4y - 6z =& 5\end{align*}\]The matrix formulation allows us to consider both the coefficients and the constant terms as part of structured mathematical operations needed for solutions.The principles of linear algebra, including operations with matrices and understanding their properties, provide foundational tools for solving even more complex systems beyond this exercise, applying to fields ranging from computational mathematics to engineering.