Problem 39 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 33-42: Use Cramer's rule, whenever applicable, to solve the system. $$ \left\\{\begin{array}{rr} x-2 y-3 z= & -1 \\ 2 x+y+z= & 6 \\ x+3 y-2 z= & 13 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{1}{5}, y = -1, z = \frac{37}{5} \).

Step 1: Write the System of Equations in Matrix Form

The given system of equations can be expressed in matrix form as \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column matrix of the variables, and \( \mathbf{b} \) is the column matrix of constants. The matrix form is:\[A = \begin{bmatrix} 1 & -2 & -3 \ 2 & 1 & 1 \ 1 & 3 & -2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -1 \ 6 \ 13 \end{bmatrix}\]

Step 2: Compute the Determinant of Matrix A

To solve the system using Cramer's Rule, first find the determinant of the coefficient matrix \( A \). The determinant is given by:\[\text{det}(A) = \begin{vmatrix} 1 & -2 & -3 \ 2 & 1 & 1 \ 1 & 3 & -2 \end{vmatrix}\]Calculate it using the formula for the determinant of a 3x3 matrix:\[\text{det}(A) = 1((1)(-2) - (1)(3)) - (-2)((2)(-2) - (1)(1)) - 3((2)(3) - (1)(1))\]Simplifying, we get:\[\text{det}(A) = -5 \]

Step 3: Solve for x Using Cramer's Rule

According to Cramer's Rule, the value of \( x \) is obtained by replacing the first column of \( A \) with \( \mathbf{b} \), and finding the determinant of this new matrix:\[A_x = \begin{bmatrix} -1 & -2 & -3 \ 6 & 1 & 1 \ 13 & 3 & -2 \end{bmatrix}\]Calculate its determinant:\[\text{det}(A_x) = (-1)((1)(-2) - (1)(3)) + (-2)((6)(-2) - (1)(13)) - 3((6)(3) - (1)(13)) = 1\]Then, \( x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{1}{-5} = -\frac{1}{5} \).

Step 4: Solve for y Using Cramer's Rule

Replace the second column of \( A \) with \( \mathbf{b} \) and find the determinant of this new matrix:\[A_y = \begin{bmatrix} 1 & -1 & -3 \ 2 & 6 & 1 \ 1 & 13 & -2 \end{bmatrix}\]Calculate its determinant:\[\text{det}(A_y) = 1((6)(-2) - (1)(13)) + 1((2)(-2) - (1)(1)) - 3((2)(13) - (1)(6)) = 5\]Then, \( y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{5}{-5} = -1 \).

Step 5: Solve for z Using Cramer's Rule

Replace the third column of \( A \) with \( \mathbf{b} \) and find the determinant of this new matrix:\[A_z = \begin{bmatrix} 1 & -2 & -1 \ 2 & 1 & 6 \ 1 & 3 & 13 \end{bmatrix}\]Calculate its determinant:\[\text{det}(A_z) = 1((1)(13) - (3)(6)) - (-2)((2)(13) - (6)(1)) - 1((2)(3) - (1)(1)) = -37\]Then, \( z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{-37}{-5} = \frac{37}{5} \).

Key Concepts

Systems of Equations

A system of equations involves multiple equations that describe the relationships between several variables. In the given exercise, we have three equations involving three variables:

• \( x - 2y - 3z = -1 \)
• \( 2x + y + z = 6 \)
• \( x + 3y - 2z = 13 \)

To solve this system, you need to find the values of \( x \), \( y \), and \( z \) that satisfy all equations simultaneously. Cramer's Rule is a handy way to solve such systems when the coefficient matrix (more on matrices later) is square (same number of equations as variables) and has a non-zero determinant. By replacing the columns of the coefficient matrix with the constants from the equations, Cramer's Rule allows you to solve for each variable separately.

Determinants

A determinant is a special number that can be calculated from a square matrix. It provides important information about the matrix, such as whether a unique solution exists for a system of equations. In our exercise, the determinant of the coefficient matrix \( A \) is calculated first:
\[ \text{det}(A) = \begin{vmatrix} 1 & -2 & -3 \ 2 & 1 & 1 \ 1 & 3 & -2 \end{vmatrix} = -5 \]
The non-zero value \(-5\) indicates that the system has a unique solution. Determinants are crucial for using Cramer's Rule because they determine the existence and uniqueness of solutions. Moreover, each variable's value is derived by determining new determinants after substituting columns of the matrix with constant terms.

Matrix Algebra

Matrix algebra is a collection of mathematical techniques that make use of matrices to handle data efficiently. It is a cornerstone for systems of equations because it simplifies computations. In this exercise, the system of equations is expressed in matrix form:
\[A = \begin{bmatrix} 1 & -2 & -3 \ 2 & 1 & 1 \ 1 & 3 & -2 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -1 \ 6 \ 13 \end{bmatrix} \]
This transformation allows us to apply operations like finding determinants or using Cramer's Rule more systematically. Matrices help in organizing and simplifying information, reducing complex systems into more manageable matrices from which we can extract valuable insights.

Linear Equations

Linear equations are mathematical expressions describing relationships where variables are to the first power and are added or subtracted together. Each equation in the exercise is linear:
\( x - 2y - 3z = -1 \), for instance, involves constants and variables only in addition, subtraction, and simple multiplication. These equations form a system that can be solved using techniques like Cramer's Rule.
Because they output a straight line when graphed, solutions to linear equations provide points of intersection that represent solutions to the system. Understanding the structure of linear equations is crucial, especially when using Cramer's Rule, which is specifically applicable to systems formed by linear equations.