Problem 22 - Algebra for College Students

Question

Use Cramer's rule to find the solution set for each system. If the equations are dependent, simply indicate that there are infinitely many solutions. \(\left(\begin{array}{rl}-5 x+6 y+4 z & =-4 \\ -7 x-8 y+2 z & =-2 \\ 2 x+9 y-z & =1\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \( x = 0 \), \( y = \frac{11}{39} \), \( z = \frac{3}{52} \).

Step 1 - Write the system in matrix form

First, express the given system of equations in the form of a matrix equation \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the column matrix of variables, and \( B \) is the column matrix of constants.\[ \begin{bmatrix} -5 & 6 & 4 \ -7 & -8 & 2 \ 2 & 9 & -1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} -4 \ -2 \ 1 \end{bmatrix} \]

Step 2 - Calculate the determinant of the coefficient matrix

Calculate the determinant of matrix \( A \) to determine if the system has a unique solution.\[ \text{det}(A) = \begin{vmatrix} -5 & 6 & 4 \ -7 & -8 & 2 \ 2 & 9 & -1 \end{vmatrix} \]Expand along the first row:\[ \text{det}(A) = -5 \begin{vmatrix} -8 & 2 \ 9 & -1 \end{vmatrix} - 6 \begin{vmatrix} -7 & 2 \ 2 & -1 \end{vmatrix} + 4 \begin{vmatrix} -7 & -8 \ 2 & 9 \end{vmatrix} \]Calculate the minors:\[ = -5((-8)(-1) - (2)(9)) - 6((-7)(-1) - (2)(2)) + 4((-7)(9) - (-8)(2)) \]\[ = -5(8 - 18) - 6(7 - 4) + 4(-63 + 16) \]\[ = -5(-10) - 6(3) + 4(-47) \]\[ = 50 - 18 - 188 = -156 \]Since \( \text{det}(A) eq 0 \), the system has a unique solution.

Step 3 - Apply Cramer's Rule to solve for x, y, and z

Using Cramer's Rule, calculate determinants of matrices formed by replacing each column of \( A \) with \( B \) one at a time to find \( x, y, \text{ and } z \).For \( x \):\[ \text{det}(A_x) = \begin{vmatrix} -4 & 6 & 4 \ -2 & -8 & 2 \ 1 & 9 & -1 \end{vmatrix} \]Calculate as before:\[ = -4 \begin{vmatrix} -8 & 2 \ 9 & -1 \end{vmatrix} - 6 \begin{vmatrix} -2 & 2 \ 1 & -1 \end{vmatrix} + 4 \begin{vmatrix} -2 & -8 \ 1 & 9 \end{vmatrix} \]\[ = -4(8 - 18) - 6(-2 + 2) + 4(-18 + 8) \]\[ = 40 - 0 - 40 = 0 \]\( x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{0}{-156} = 0 \)For \( y \): \[ \text{det}(A_y) = \begin{vmatrix} -5 & -4 & 4 \ -7 & -2 & 2 \ 2 & 1 & -1 \end{vmatrix} \]Calculate similarly:\[ = -5(-2(-1) - 1(2)) + 4(-7(1) - 2(2)) \]\[ = -5(2 - 2) + 4(-7 - 4) \]\[ = 0 + 4(-11) = -44 \]\( y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-44}{-156} = \frac{11}{39} \)For \( z \):\[ \text{det}(A_z) = \begin{vmatrix} -5 & 6 & -4 \ -7 & -8 & -2 \ 2 & 9 & 1 \end{vmatrix} \]Calculate:\[ = -5(-8(1) - 9(-2)) - 6(-7(1) - 2(2)) \]\[ = -5(15) + 6(11) \]\[ = -75 + 66 = -9 \]\( z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{-9}{-156} = \frac{3}{52} \)

Key Concepts

Determinants

Determinants play a crucial role in understanding many concepts in linear algebra, especially when using Cramer's Rule. A determinant provides a single number that can be calculated from a square matrix, and it helps in understanding the properties of the matrix.

• If the determinant of a matrix is zero, the matrix is said to be singular, which means no unique solutions exist for equations using this matrix.
• If the determinant is not zero, the system of equations will typically have a unique solution.

To calculate the determinant of a 3x3 matrix, such as in our exercise, we first select a row or a column and expand it using the method of cofactors. This involves selecting an element, multiplying it by the determinant of a smaller 2x2 matrix derived from the original matrix, and then alternating the addition and subtraction of products. This expansion helps in determining whether systems of equations will have unique solutions, multiple solutions, or none.

Matrix Equations

Matrix equations are a compact way of expressing systems of linear equations. Instead of writing out each individual equation, we represent them as matrices — a powerful tool that aids in solving these systems efficiently.

• Each equation in a system can be represented as a row in a matrix.
• The variables of the equations form a column matrix, and the solutions form another column matrix.

In the exercise, the system of equations is given by a matrix equation in the form of \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix as \( \begin{bmatrix} x \ y \ z \end{bmatrix} \), and \( B \) is the constant matrix. Solving the matrix equation involves finding \( X \). This can be achieved using methods like matrix inversion or, as in this exercise, by applying Cramer's Rule.

Systems of Equations

Systems of equations are sets of equations with multiple variables, where each equation shares the same set of variables. The goal here is to find values of these variables that satisfy all equations simultaneously.
There are different types of solutions a system can have:

• Unique solution: Only one set of values satisfies all equations.
• Infinite solutions: Every equation describes the same line, plane, or hyperplane, so there are multiple solutions lying along this line or plane.
• No solution: Equations describe parallel lines or planes that never meet.

In our example, the determinant is not zero, indicating a unique solution for the system. Cramer's Rule provides a structured approach for calculating each variable's value, showing that a consistent set of unique solutions exists.

Algebra

Algebra is a branch of mathematics dealing with symbols and rules for manipulating those symbols. In our context, it refers to solving linear algebraic equations using different methods and rules, such as Cramer's Rule.
Using algebraic operations, systems of equations can be solved to find unknown values.

• Substitution involves solving one equation for one variable and substituting this back into the other equations.
• Elimination involves adding or subtracting equations to eliminate a variable.

Cramer's Rule is an algebraic technique that is especially useful for systems where the number of equations matches the number of unknowns, and it provides a neat formulaic method to arrive at the solution quickly without lengthy algebraic manipulations. It leverages determinants and matrix algebra to find each variable's value in an elegant and compact form. This makes it a favourite in theoretical and practical applications.