Problem 71 - A Graphical Approach to Precalculus with Limits

Question

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}2 x-y+3 z &=1 \\\\-2 x+y-3 z &=2 \\\5 x-y+z &=2\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = -1.5 \), \( y = 4 \), \( z = 3 \).

Step 1: Write the system in matrix form

The system of equations can be written in matrix form as \( AX = B \), where \( A \) is the coefficient matrix:\[A = \begin{bmatrix} 2 & -1 & 3 \ -2 & 1 & -3 \ 5 & -1 & 1 \end{bmatrix}\]and \( B \) is the constant matrix:\[B = \begin{bmatrix} 1 \ 2 \ 2 \end{bmatrix}\]The solution vector \( X \) is \( \begin{bmatrix} x \ y \ z \end{bmatrix} \).

Step 2: Calculate the determinant of matrix A

To use Cramer's rule, first calculate the determinant of the coefficient matrix \( A \):\[D = \begin{vmatrix} 2 & -1 & 3 \ -2 & 1 & -3 \ 5 & -1 & 1 \end{vmatrix}\]Using the rule of Sarrus or cofactor expansion, compute the determinant:\[D = 2(1 \cdot 1 - (-1) \cdot (-3)) - (-1)(-2 \cdot 1 - 5 \cdot (-3)) + 3(-2 \cdot (-1) - 5 \cdot 1)\]\[D = 2(1 - 3) + 1(2 + 15) + 3(2 - 5)\]\[D = 2(-2) + 1(17) + 3(-3)\]\[D = -4 + 17 - 9 = 4\]

Step 3: Check if D = 0

Since \( D eq 0 \) (specifically \( D = 4 \)), Cramer's rule can be applied to solve the system of equations directly.

Step 4: Calculate determinants for numerator matrices

To find \( x, y, \) and \( z \), replace the respective column of \( A \) with matrix \( B \) and calculate the determinant:\( D_x = \begin{vmatrix} 1 & -1 & 3 \ 2 & 1 & -3 \ 2 & -1 & 1 \end{vmatrix} \), \( D_y = \begin{vmatrix} 2 & 1 & 3 \ -2 & 2 & -3 \ 5 & 2 & 1 \end{vmatrix} \), \( D_z = \begin{vmatrix} 2 & -1 & 1 \ -2 & 1 & 2 \ 5 & -1 & 2 \end{vmatrix} \).

Step 5: Calculate determinant D_x

Evaluate \( D_x \):\[D_x = 1(1 \cdot 1 - (-1) \cdot (-3)) - (-1)(2 \cdot 1 - 2 \cdot (-3)) + 3(2 \cdot (-1) - 2 \cdot 1)\]\[D_x = 1(1 - 3) + 1(2 + 6) + 3(-2 - 2)\]\[D_x = -2 + 8 - 12 = -6\]

Step 6: Calculate determinant D_y

Evaluate \( D_y \):\[D_y = 2(2 \cdot 1 - (-1) \cdot 1) - 1(-2 \cdot 1 - 5 \cdot 1) + 3(-2 \cdot 2 - 5 \cdot (-1))\]\[D_y = 2(2 + 1) + 1(2 + 5) + 3(-4 + 5)\]\[D_y = 6 + 7 + 3 = 16\]

Step 7: Calculate determinant D_z

Evaluate \( D_z \):\[D_z = 2(1 \cdot 2 - (-1) \cdot 1) - (-1)(-2 \cdot 2 - 5 \cdot 1) + 1(-2 \cdot (-1) - 5 \cdot 1)\]\[D_z = 2(2 + 1) + 1(4 + 5) + 1(2 - 5)\]\[D_z = 6 + 9 - 3 = 12\]

Step 8: Apply Cramer's Rule

Using Cramer's rule, calculate the variables:\[x = \frac{D_x}{D} = \frac{-6}{4} = -1.5\]\[y = \frac{D_y}{D} = \frac{16}{4} = 4\]\[z = \frac{D_z}{D} = \frac{12}{4} = 3\]

Key Concepts

System of Equations

A system of equations is a collection of two or more equations that share the same set of unknowns. In our example, we have a system with three equations and three unknowns: \( x \), \( y \), and \( z \). The goal is to find values for these variables that satisfy all the equations simultaneously. Typically, systems of equations can be represented in various forms, such as:

• Standard form: where equations are written separately.
• Matrix form: using matrices to represent the system.

In this exercise, the given system of equations is: \[\begin{aligned}2x - y + 3z &= 1 \-2x + y - 3z &= 2 \5x - y + z &= 2\end{aligned}\]The beauty of using a consistent method like Cramer's Rule lies in its simplicity when handling linear equations, especially when the number of equations matches the number of unknowns.

Determinant

Determinants play a crucial role in solving systems of linear equations using matrix algebra. A determinant is a scalar value derived from a square matrix and provides vital information about the matrix itself, such as whether a unique solution exists. For a 3x3 matrix, the determinant is calculated using a specific formula, often utilizing cofactor expansion. In our exercise, the determinant of the coefficient matrix \( A \) is calculated as follows: \[D = \begin{vmatrix} 2 & -1 & 3 \-2 & 1 & -3 \5 & -1 & 1 \end{vmatrix}\]By applying the rule of Sarrus or cofactor expansion, the value of the determinant \( D \) is found to be 4, indicating that the system is consistent and a unique solution can be achieved using Cramer's Rule. In summary, the determinant is key to deciding if Cramer's Rule can be applied. If \( D = 0 \), it would suggest the system might have no solution or infinitely many solutions, requiring alternative methods for solving.

Matrix Algebra

Matrix algebra is a mathematical tool used for handling and solving systems of equations efficiently. In matrix form, a system of linear equations can be expressed as \( AX = B \), where:

• \( A \) is the coefficient matrix holding the coefficients of the variables.
• \( X \) is the solution vector with the variables \( x, y, z \).
• \( B \) is the constant matrix containing the constants from the equations.

In our specific case, the matrices are:\[A = \begin{bmatrix}2 & -1 & 3 \-2 & 1 & -3 \5 & -1 & 1\end{bmatrix}, \quadB = \begin{bmatrix}1 \2 \2\end{bmatrix}, \quadX = \begin{bmatrix}x \y \z\end{bmatrix}\]Using matrices allows for a structured and efficient computational approach, especially when leveraging determinant-related methods like Cramer's Rule. It simplifies the equation handling process by allowing simultaneous operations on all components of the system.

Solution Method

Cramer's Rule is a direct method used for solving systems of linear equations, provided the determinant of the coefficient matrix is non-zero. This method involves replacing each column of the coefficient matrix \( A \) with the constant matrix \( B \) to form new matrices, and then calculating their determinants to find the variables. Here's how it works in our example:

• Calculate \( D \), the determinant of the original coefficient matrix.
• Form matrix \( A_x \) by replacing the first column of \( A \) with \( B \), and calculate \( D_x \).
• Form matrix \( A_y \) by replacing the second column with \( B \), and calculate \( D_y \).
• Form matrix \( A_z \) by replacing the third column with \( B \), and calculate \( D_z \).

The solutions for \( x, y, \) and \( z \) are then determined using: \[x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}\] For our system, the calculations provide \( x = -1.5 \), \( y = 4 \), and \( z = 3 \). This rule is effective and straightforward but requires that the determinant is non-zero. If \( D = 0 \), other methods, such as Gaussian elimination, might be needed to find a solution or determine the specific nature of the system.