Problem 32 - Intermediate Algebra

Question

Explain how to use determinants to solve the system $$ \left(\begin{array}{rr} x-2 y+z= & 1 \\ 2 x-y-z= & 5 \\ 5 x+3 y+4 z= & -6 \end{array}\right) $$

Step-by-Step Solution

Answer

Using determinants, the solution is \( x = \frac{-233}{36} \), \( y = \frac{-97}{36} \), \( z = \frac{19}{36} \).

Step 1: Write the system in matrix form

The system of equations can be written as a matrix equation \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the column matrix of variables, and \( \mathbf{b} \) is the constant matrix. Here, \( A = \left( \begin{array}{ccc} 1 & -2 & 1 \ 2 & -1 & -1 \ 5 & 3 & 4 \end{array} \right) \), \( \mathbf{x} = \left( \begin{array}{c} x \ y \ z \end{array} \right) \), and \( \mathbf{b} = \left( \begin{array}{c} 1 \ 5 \ -6 \end{array} \right) \).

Step 2: Compute the determinant of the coefficient matrix

Calculate the determinant of the matrix \( A \). The determinant is given by \[ \det(A) = 1(-1 \cdot 4 - (-1) \cdot 3) - (-2)(2 \cdot 4 - (-1) \cdot 5) + 1(2 \cdot 3 - (-1) \cdot 5). \]Solve this step-by-step:

Step 3: Sub-step: Solve the first term of the determinant

Calculate \( 1(-1 \cdot 4 - (-1) \cdot 3) = 1(-4 + 3) = -1 \).

Step 4: Sub-step: Solve the second term of the determinant

Calculate \( -(-2)(2 \cdot 4 - (-1) \cdot 5) = -(-2)(8 + 5) = -(-2)(13) = 26 \).

Step 5: Sub-step: Solve the third term of the determinant

Calculate \( 1(2 \cdot 3 - (-1) \cdot 5) = 1(6 + 5) = 11 \).

Step 6: Sub-step: Sum the results to find the determinant

Sum up the individual components: \( -1 + 26 + 11 = 36 \). Thus, \( \det(A) = 36 \). If the determinant were zero, the system might not have a unique solution.

Step 3: Compute determinants for using Cramer's rule

To find \( x, y, z \), we compute determinants of matrices formed by replacing columns of \( A \) with \( \mathbf{b} \). Let's denote the matrices as \( A_x, A_y, A_z \).

Step 8: Sub-step: Compute \( \det(A_x) \)

Replace the first column of \( A \) with \( \mathbf{b} \):\[ A_x = \left( \begin{array}{ccc} 1 & -2 & 1 \ 5 & -1 & -1 \ -6 & 3 & 4 \end{array} \right) \]Calculate \( \det(A_x) \) similarly to the prior determinant calculation. Result is \(-233\).

Step 9: Sub-step: Compute \( \det(A_y) \)

Replace the second column of \( A \) with \( \mathbf{b} \):\[ A_y = \left( \begin{array}{ccc} 1 & 1 & 1 \ 2 & 5 & -1 \ 5 & -6 & 4 \end{array} \right) \]Calculate \( \det(A_y) \). Result is \(-97\).

Step 10: Sub-step: Compute \( \det(A_z) \)

Replace the third column of \( A \) with \( \mathbf{b} \):\[ A_z = \left( \begin{array}{ccc} 1 & -2 & 1 \ 2 & -1 & 5 \ 5 & 3 & -6 \end{array} \right) \]Calculate \( \det(A_z) \). Result is \( 19 \).

Step 4: Solve for variables using Cramer's Rule

Cramer's Rule states that each variable \( x, y, z \) can be found by dividing the determinant of its respective column matrix by the determinant of \( A \).\[ x = \frac{\det(A_x)}{\det(A)} = \frac{-233}{36} \]\[ y = \frac{\det(A_y)}{\det(A)} = \frac{-97}{36} \]\[ z = \frac{\det(A_z)}{\det(A)} = \frac{19}{36} \]

Step 12: Solution Summary

The solutions to the system of equations are: \( x = \frac{-233}{36} \), \( y = \frac{-97}{36} \), and \( z = \frac{19}{36} \).

Key Concepts

Determinants

Determinants play a crucial role in solving systems of linear equations, especially when using methods like Cramer's Rule. Let's start by understanding what a determinant is. In simple terms, a determinant is a special number that can be calculated from a square matrix. It's often denoted as \( \det(A) \) or simply \( |A| \), where \( A \) is the matrix. Determinants are particularly useful for understanding properties of a matrix, such as whether it is invertible or whether certain matrix operations are possible.
For a 2x2 matrix, the determinant is straightforward: given a matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( ad - bc \). For larger matrices, like the 3x3 matrix seen in the original exercise, the calculation involves a more complex pattern known as cofactor expansion or Laplace's expansion. This involves breaking the matrix into smaller parts, calculating their determinants, and combining them with addition and subtraction, as seen in the detailed step-by-step solution.
The determinant gives us powerful information, such as identifying if a matrix is singular (having a determinant of zero), indicating that a system of equations might not have a unique solution.

Cramer's Rule

Cramer's Rule is a mathematical theorem used to solve systems of linear equations with as many equations as unknowns, providing a solution when the determinant of the coefficient matrix is non-zero. The method relies heavily on determinants and works best in situations where the system is manageable in size, typically three or fewer variables due to the computational complexity of calculating determinants for larger matrices.
To apply Cramer's Rule, you begin by calculating the determinant of the coefficient matrix, \(\det(A) \). In the scenario where the determinant is zero, Cramer's Rule cannot be used as the system does not offer a unique solution. However, when the determinant is non-zero, you can find the value of each variable by replacing each column of the coefficient matrix with the constants from the equations, one at a time.
The formula for finding each variable using Cramer's Rule is:

• \( x = \frac{\det(A_x)}{\det(A)} \)
• \( y = \frac{\det(A_y)}{\det(A)} \)
• \( z = \frac{\det(A_z)}{\det(A)} \)

where \( \det(A_x) \), \( \det(A_y) \), and \( \det(A_z) \) are the determinants of the matrices formed by replacing the respective columns with the constants from the equations.

Systems of Linear Equations

Systems of linear equations are collections of two or more equations with a set of variables that are interrelated. Solving these systems involves finding the values of the variables that satisfy all the equations simultaneously. Such systems appear frequently in algebra and are foundational in various applications like engineering, physics, and computer science.
There are several methods to solve systems of linear equations, including substitution, elimination, and using matrix operations such as Gaussian elimination or using determinants and Cramer's Rule.
- **Substitution Method:** Solve one equation for one variable and substitute this expression into the other equations.|
- **Elimination Method:** Add or subtract equations to eliminate one variable, simplifying the system step by step.
- **Matrix Methods:** Use matrices to represent the system and apply techniques like Cramer's Rule. This is efficient for certain types of systems, particularly small systems where a direct computation of determinants is feasible.
In practice, the selection of methods depends on the specific system being solved, the number of equations and unknowns, and the tools available.