Problem 40 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{array}{r} 5 x+2 y-z=1 \\ -7 x-8 y+3 z=1.5 \\ 6 x-12 y+z=7 \end{array} $$

Step-by-Step Solution

Answer

(x, y, z) = (6.4375, 6.3125, 18.9375)

Step 1: Write the System of Equations in Matrix Form

Represent the given system of equations with matrices. The system is\[ \begin{pmatrix} 5 & 2 & -1 \ -7 & -8 & 3 \ 6 & -12 & 1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ 1.5 \ 7 \end{pmatrix} \] where \( A \) is the matrix of coefficients, \( \mathbf{x} \) is the column matrix of variables \([x, y, z]^T\), and \( \mathbf{b} \) is the column matrix of constants.

Step 2: Compute the Determinant of Matrix A (det(A))

Calculate the determinant of the coefficient matrix \( A \):\[A = \begin{pmatrix} 5 & 2 & -1 \ -7 & -8 & 3 \ 6 & -12 & 1 \end{pmatrix}\]Using the formula for the determinant of a 3x3 matrix, we find:\[det(A) = 5((-8)(1) - (3)(-12)) + 2((-7)(1) - (3)(6)) + (-1)((-7)(-12) - (-8)(6))\]Calculate each term to find:\[det(A) = 5(28) + 2(-60) - 1(36) = 140 - 120 - 36 = -16\]Thus, \( det(A) = -16 \).

Step 3: Create Matrices A_x, A_y, and A_z

To use Cramer's Rule, create three matrices by replacing one column of \( A \) with \( \mathbf{b} \):- \( A_x \) is \( A \) with the first column replaced by \( \mathbf{b} \).\[A_x = \begin{pmatrix} 1 & 2 & -1 \ 1.5 & -8 & 3 \ 7 & -12 & 1 \end{pmatrix}\]- \( A_y \) is \( A \) with the second column replaced by \( \mathbf{b} \).\[A_y = \begin{pmatrix} 5 & 1 & -1 \ -7 & 1.5 & 3 \ 6 & 7 & 1 \end{pmatrix}\]- \( A_z \) is \( A \) with the third column replaced by \( \mathbf{b} \).\[A_z = \begin{pmatrix} 5 & 2 & 1 \ -7 & -8 & 1.5 \ 6 & -12 & 7 \end{pmatrix}\]

Step 4: Compute Determinants det(A_x), det(A_y), and det(A_z)

Calculate the determinants for \( A_x \), \( A_y \), and \( A_z \).For \( det(A_x) \):\[det(A_x) = 1(-8 \cdot 1 - 3 \cdot -12) + 2(1.5 \cdot 1 - 3 \cdot 7) - 1(-7 \cdot -12 - (-8 \cdot 6))\]\[det(A_x) = 44 - 39 - 84 = -103\]For \( det(A_y) \):\[det(A_y) = 5(1.5 \cdot 1 - 3 \cdot 7) - 1(5 \cdot 1 - 3 \cdot 6) - 1.5(-7 \cdot 1 - 1 \cdot 6)\]\[det(A_y) = -91.5 - (5 - 18) + 9.5 = -101\]For \( det(A_z) \):\[det(A_z) = 5(-8 \cdot 7 - 1.5 \cdot -12) + 2(-7 \cdot 7 - 1 \cdot 6) + 1(7(-12) - -8 \cdot 6)\]\[det(A_z) = -305 - 110 + 112 = -303\]

Step 5: Apply Cramer's Rule to Find the Solution

Using Cramer's Rule, each variable is found by \( x = \frac{det(A_x)}{det(A)} \), \( y = \frac{det(A_y)}{det(A)} \), \( z = \frac{det(A_z)}{det(A)} \).Thus:- \( x = \frac{-103}{-16} = 6.4375 \)- \( y = \frac{-101}{-16} = 6.3125 \)- \( z = \frac{-303}{-16} = 18.9375 \)The solution to the system of equations is \((x, y, z) = (6.4375, 6.3125, 18.9375)\).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations involving the same set of variables. These equations represent lines that can intersect at a point, run parallel, or be the same line.

The goal in solving a system of linear equations is to find the values of the variables that satisfy all given equations simultaneously.

In the context of a 3-variable system, we typically have equations that are expressed in the form:

\[ ax + by + cz = d \]

where \( a, b, \) and \( c \) are coefficients, \( x, y, \) and \( z \) are variables, and \( d \) is a constant. This is exactly what we see in the problem, which consists of three such equations.

To solve these using Cramer's Rule, the system is expressed in matrix form where coefficients of variables form a square matrix. This approach is especially useful when the system is 3x3, as it leverages properties of determinants.

Determinant Calculation

The determinant is a scalar value that can be computed from the elements of a square matrix and encapsulates important properties of the matrix. It is helpful in determining whether a system of equations has a unique solution.

For a 3x3 matrix, the determinant gives essential insights into the relationships among the equations. If the determinant is non-zero, the system has a unique solution.

The formula for finding the determinant of a 3x3 matrix defined as \( A = \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \) is:

\[ det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

In our problem, we used this formula to find \( det(A) = -16 \), ensuring the system can be solved using Cramer's Rule, as the determinant is not zero.

3x3 Matrix

A 3x3 matrix in the context of solving linear equations involves three rows and three columns of coefficients. This matrix represents the left-hand side of a system of three linear equations.

In Cramer's Rule, the original 3x3 matrix, often denoted as \( A \), forms the core structure on which we build further calculations. Modifications to this matrix, such as replacing columns with the constant terms of the equations, lead to variants \( A_x, A_y, \) and \( A_z \) used to solve for individual variables.

Each of these matrices retains the general structure of \( A \) but with one column swapped with the constants column \( \ b \).

This process emphasizes why understanding and manipulating 3x3 matrices is crucial in linear algebra, as it builds the foundation for finding solutions to systems using determinant-based methods like Cramer's Rule.