Question

Solve each system of equations using Cramer's Rule if is applicable. If Cramer's Rule is not applicable, write, "Not applicable" \(\left\\{\begin{array}{r}x+y-z=6 \\ 3 x-2 y+z=-5 \\ x+3 y-2 z=14\end{array}\right.\)

Short answer

The solution is \( x = 1 \), \( y = 3 \), and \( z = -2 \).

Step-by-step solution

- Write the System in Matrix Form

First, represent the system of equations in matrix form. The system can be written as: \[ A \mathbf{x} = \mathbf{b} \] where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constants matrix. \[ A = \begin{bmatrix} 1 & 1 & -1 \ 3 & -2 & 1 \ 1 & 3 & -2 \end{bmatrix} \] \[ \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \] \[ \mathbf{b} = \begin{bmatrix} 6 \ -5 \ 14 \end{bmatrix} \]

- Compute the Determinant of Matrix A

Calculate the determinant of the coefficient matrix \( A \). The determinant, \( \Delta \), is given by: \[ \Delta = \text{det}(A) = \begin{vmatrix} 1 & 1 & -1 \ 3 & -2 & 1 \ 1 & 3 & -2 \end{vmatrix} \] Using the cofactor expansion, we find: \[ \Delta = 1 \cdot \begin{vmatrix} -2 & 1 \ 3 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 1 \ 1 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -2 \ 1 & 3 \end{vmatrix} \] Simplifying further:\[ \Delta = 1((-2)(-2) - (1)(3)) - 1((3)(-2) - (1)(1)) - 1((3)(3) - (-2)(1)) \] \[ \Delta = 1(4 - 3) - 1(-6 - 1) - 1(9 + 2) \] \[ \Delta = 1 - (-7) - 11 \] \[ \Delta = 1 + 7 - 11 = -3 \]

- Compute the Determinant for Matrix Dx

Replace the first column of \( A \) with \( \mathbf{b} \) to form \( D_x \). Then compute the determinant of \( D_x \): \[ D_x = \begin{vmatrix} 6 & 1 & -1 \ -5 & -2 & 1 \ 14 & 3 & -2 \end{vmatrix} \] Using the cofactor expansion, we find: \[ D_x = 6 \cdot \begin{vmatrix} -2 & 1 \ 3 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} -5 & 1 \ 14 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} -5 & -2 \ 14 & 3 \end{vmatrix} \] Simplifying:\[ D_x = 6((-2)(-2) - (1)(3)) - 1((-5)(-2) - (1)(14)) - 1((-5)(3) - (-2)(14)) \] \[ D_x = 6(4 - 3) - 1(10 - 14) - 1(-15 + 28) \] \[ D_x = 6(1) - 1(-4) - 1(13) \] \[ D_x = 6 + 4 - 13 = -3 \]

- Compute the Determinant for Matrix Dy

Replace the second column of \( A \) with \( \mathbf{b} \) to form \( D_y \). Then compute the determinant of \( D_y \): \[ D_y = \begin{vmatrix} 1 & 6 & -1 \ 3 & -5 & 1 \ 1 & 14 & -2 \end{vmatrix} \] Using the cofactor expansion, we get:\[ D_y = 1 \cdot \begin{vmatrix} -5 & 1 \ 14 & -2 \end{vmatrix} - 6 \cdot \begin{vmatrix} 3 & 1 \ 1 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -5 \ 1 & 14 \end{vmatrix} \] Simplifying further:\[ D_y = 1((-5)(-2) - (1)(14)) - 6((3)(-2) - (1)(1)) - 1((3)(14) - (-5)(1)) \] \[ D_y = 1(10 - 14) - 6(-6 - 1) - 1(42 + 5) \] \[ D_y = 1(-4) - 6(-7) - 47 \] \[ D_y = -4 + 42 - 47 = -9 \]

- Compute the Determinant for Matrix Dz

Replace the third column of \( A \) with \( \mathbf{b} \) to form \( D_z \). Then compute the determinant of \( D_z \): \[ D_z = \begin{vmatrix} 1 & 1 & 6 \ 3 & -2 & -5 \ 1 & 3 & 14 \end{vmatrix} \] Using the cofactor expansion, we find:\[ D_z = 1 \cdot \begin{vmatrix} -2 & -5 \ 3 & 14 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -5 \ 1 & 14 \end{vmatrix} + 6 \cdot \begin{vmatrix} 3 & -2 \ 1 & 3 \end{vmatrix} \] Simplifying:\[ D_z = 1((-2)(14) - (-5)(3)) - 1((3)(14) - (-5)(1)) + 6((3)(3) - (-2)(1)) \] \[ D_z = 1(-28 + 15) - 1(42 + 5) + 6(9 + 2) \] \[ D_z = -13 - 47 + 66 = 6 \]

- Solve for Variables using Cramer's Rule

Using Cramer's Rule, solve for \( x \), \( y \), and \( z \). The formulas are: \[ x = \frac{D_x}{\Delta} = \frac{-3}{-3} = 1 \] \[ y = \frac{D_y}{\Delta} = \frac{-9}{-3} = 3 \] \[ z = \frac{D_z}{\Delta} = \frac{6}{-3} = -2 \]Thus, the solution to the system is: \( x = 1 \), \( y = 3 \), and \( z = -2 \).

Key concepts

System of Equations

A system of equations is a set of two or more equations that have the same set of variables. In this exercise, we have three equations with three unknowns: \( x \), \( y \), and \( z \). The goal is to find a common solution that satisfies all equations simultaneously. We use Cramer's Rule in this example, which relies on the properties of determinants from linear algebra.
A system of linear equations can be represented in matrix form as \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the matrix of coefficients, \( \mathbf{x} \) is the column matrix of variables, and \( \mathbf{b} \) is the constants column matrix. Breaking down the system into matrices helps in applying various mathematical techniques for solving the system.
Determining whether a system has a unique solution, infinitely many solutions, or no solution at all is essential. For Cramer's Rule to be applicable, the determinant of the matrix \( A \) must be nonzero (\( \Delta e 0 \)). In this problem, we have: \( \mathbf{x} \=\begin{bmatrix}x \ y\ z\end{bmatrix} \), \( \mathbf{b} \=\begin{bmatrix} 6 \ -5 \ 14 \end{bmatrix} \). Notice that each element of the matrices corresponds to coefficients and constants from the original system of equations.

Matrix Determinants

Determinants are scalar values that provide important information about a matrix. They are particularly useful in solving systems of linear equations. For a 3x3 matrix, the determinant is calculated using the rule of Sarrus or cofactor expansion.
For example, the determinant of matrix \( A \): \( A = \begin{bmatrix} 1 & 1 & -1 \ 3 & -2 & 1 \ 1 & 3 & -2 \end{bmatrix} \) is computed as follows: \[ \Delta = 1 \cdot \begin{vmatrix} -2 & 1 \ 3 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 1 \ 1 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -2 \ 1 & 3 \end{vmatrix}\] Using cofactor expansion, we simplify it to: \[-3\]. Determinants help determine if a system of equations has a unique solution (if \( \Delta e 0 \)), no solution, or infinitely many solutions (if \( \Delta = 0 \)).
The principle of replacing columns with the constants matrix to find specific determinants, as done here with \( \[ D_x, D_y, D_z \] \), is a key step in Cramer's Rule.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It plays a vital role in various fields such as engineering, physics, computer science, and economics.
In solving the given system of equations using Cramer's Rule, we leveraged several core concepts of linear algebra, including matrix representation, determinant calculation, and matrix manipulation. Understanding these fundamentals allows us to solve complex problems systematically.
Key linear algebra concepts include:

• Vector spaces
• Linear transformations
• Matrix operations (addition, multiplication, inversion)
• Eigenvalues and eigenvectors

In this specific exercise, understanding how to manipulate matrices and compute their determinants is crucial for applying Cramer's Rule effectively. By mastering these tools, one can tackle various problems involving systems of equations and beyond.