Question

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

Short answer

x = -2, y = -\frac{1}{12}, z = -1

Step-by-step solution

Write the system of equations in matrix form

To apply Cramer's Rule, express the system of equations in matrix form: \[ A \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -2 \ -5 \ 5 \end{pmatrix} \] Where matrix A is the coefficient matrix: \[ A = \begin{pmatrix} 1 & 3 & -1 \ 2 & -6 & 1 \ -3 & 3 & -2 \end{pmatrix} \]

Calculate the determinant of matrix A

Find the determinant of the coefficient matrix, \( \text{det}(A) \): \[ \text{det}(A) = \begin{vmatrix} 1 & 3 & -1 \ 2 & -6 & 1 \ -3 & 3 & -2 \end{vmatrix} \] Use the cofactor expansion method: \[ \text{det}(A) = 1 \begin{vmatrix} -6 & 1 \ 3 & -2 \end{vmatrix} - 3 \begin{vmatrix} 2 & 1 \ -3 & -2 \end{vmatrix} - 1 \begin{vmatrix} 2 & -6 \ -3 & 3 \end{vmatrix} \] \[ \text{det}(A) = 1((-6)(-2) - (1)(3)) - 3((2)(-2) - (1)(-3)) - 1((2)(3) - (-6)(-3)) \] \[ \text{det}(A) = 1(12 - 3) - 3(-4 + 3) - 1(6 - 18) \] \[ \text{det}(A) = 1(9) - 3(-1) - 1(-12) \] \[ \text{det}(A) = 9 + 3 + 12 = 24 \]

Determine if Cramer's Rule is applicable

Since \( \text{det}(A) eq 0 \) (it is 24), Cramer's Rule is applicable. Now, calculate the determinants needed for the numerators in Cramer's Rule.

Calculate determinant for x (\text{det}(A_x))

Create matrix \( A_x \) by replacing the first column of \( A \) with the constants: \[ A_x = \begin{pmatrix} -2 & 3 & -1 \ -5 & -6 & 1 \ 5 & 3 & -2 \end{pmatrix} \] Calculate \(\text{det}(A_x)\): \[ \text{det}(A_x) = \begin{vmatrix} -2 & 3 & -1 \ -5 & -6 & 1 \ 5 & 3 & -2 \end{vmatrix} \] \[ \text{det}(A_x) = -2 \begin{vmatrix} -6 & 1 \ 3 & -2 \end{vmatrix} - 3 \begin{vmatrix} -5 & 1 \ 5 & -2 \end{vmatrix} - 1 \begin{vmatrix} -5 & -6 \ 5 & 3 \end{vmatrix} \] \[ \text{det}(A_x) = -2((-6)(-2) - (1)(3)) - 3((-5)(-2) - (1)(5)) - 1((-5)(3) - (-6)(5)) \] \[ \text{det}(A_x) = -2(12 - 3) - 3(10 - 5) - 1(-15 + 30) \] \[ \text{det}(A_x) = -2(9) - 3(5) - 1(15) \] \[ \text{det}(A_x) = -18 - 15 - 15 = -48 \]

Calculate determinant for y (\text{det}(A_y))

Create matrix \( A_y \) by replacing the second column of \( A \) with the constants: \[ A_y = \begin{pmatrix} 1 & -2 & -1 \ 2 & -5 & 1 \ -3 & 5 & -2 \end{pmatrix} \] Calculate \(\text{det}(A_y)\): \[ \text{det}(A_y) = \begin{vmatrix} 1 & -2 & -1 \ 2 & -5 & 1 \ -3 & 5 & -2 \end{vmatrix} \] \[ \text{det}(A_y) = 1 \begin{vmatrix} -5 & 1 \ 5 & -2 \end{vmatrix} - (-2) \begin{vmatrix} 2 & 1 \ -3 & -2 \end{vmatrix} - (-1) \begin{vmatrix} 2 & -5 \ -3 & 5 \end{vmatrix} \] \[ \text{det}(A_y) = 1((-5)(-2) - (1)(5)) - (-2)((2)(-2) - (1)(-3)) - (-1)((2)(5) - (-5)(-3)) \] \[ \text{det}(A_y) = 1(10 - 5) + 2(-4 + 3) + 1(10 - 15) \] \[ \text{det}(A_y) = 5 + 2(-1) + 1(-5) \] \[ \text{det}(A_y) = 5 - 2 - 5 = -2 \]

Calculate determinant for z (\text{det}(A_z))

Create matrix \( A_z \) by replacing the third column of \( A \) with the constants: \[ A_z = \begin{pmatrix} 1 & 3 & -2 \ 2 & -6 & -5 \ -3 & 3 & 5 \end{pmatrix} \] Calculate \(\text{det}(A_z)\): \[ \text{det}(A_z) = \begin{vmatrix} 1 & 3 & -2 \ 2 & -6 & -5 \ -3 & 3 & 5 \end{vmatrix} \] \[ \text{det}(A_z) = 1 \begin{vmatrix} -6 & -5 \ 3 & 5 \end{vmatrix} - 3 \begin{vmatrix} 2 & -5 \ -3 & 5 \end{vmatrix} - (-2) \begin{vmatrix} 2 & -6 \ -3 & 3 \end{vmatrix} \] \[ \text{det}(A_z) = 1((-6)(5) - (-5)(3)) - 3((2)(5) - (-5)(-3)) + 2((2)(3) - (-6)(-3)) \] \[ \text{det}(A_z) = 1(-30 + 15) - 3(10 - 15) + 2(6 - 18) \] \[ \text{det}(A_z) = 1(-15) - 3(-5) + 2(-12) \] \[ \text{det}(A_z) = -15 + 15 - 24 = -24 \]

Find the values of x, y, and z

Use the determinants to find the values of \(x\), \(y\), and \(z\) using Cramer's Rule: \[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-48}{24} = -2 \] \[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-2}{24} = -\frac{1}{12} \] \[ z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{-24}{24} = -1 \]

Key concepts

system of equations

A system of equations is a set of two or more equations with the same variables. These equations are solved simultaneously to find a common solution for the variables. In the exercise provided, our system of equations consists of three equations:

\[x + 3y - z = -2\]
\[2x - 6y + z = -5\]
\[-3x + 3y - 2z = 5\]
To solve this system, we'll use a method called Cramer's Rule. This involves converting the system into matrix form and working with determinants to find the values of the variables.

determinants

Determinants are a key concept in matrix algebra and linear algebra. They are a scalar value that can be computed from the elements of a square matrix. The determinant helps in finding the solutions to a system of linear equations, among other applications.

For our example, the determinant of the 3x3 coefficient matrix \( A \) is calculated using the cofactor expansion method:
\[ \text{det}(A) = \begin{pmatrix} 1 & 3 & -1 \ 2 & -6 & 1 \ -3 & 3 & -2 \end{pmatrix} \]
We expand along the first row:
\[ \text{det}(A) = 1 \begin{vmatrix} -6 & 1 \ 3 & -2 \end{vmatrix} - 3 \begin{vmatrix} 2 & 1 \ -3 & -2 \end{vmatrix} - 1 \begin{vmatrix} 2 & -6 \ -3 & 3 \end{vmatrix} \]
Which results in:
\[ \text{det}(A) = 9 + 3 + 12 = 24 \]
Since the determinant is non-zero, Cramer's Rule is applicable.

matrix algebra

Matrix algebra involves operations with matrices: addition, subtraction, multiplication, and finding the determinant, among others. In this exercise, we'll focus primarily on writing the system of equations in matrix form and computing determinants.

Our system of equations is represented in matrix form as:
\[ A \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} -2 \ -5 \ 5 \end{pmatrix} \]
Where \( A \) is the coefficient matrix:
\[ A = \begin{pmatrix} 1 & 3 & -1 \ 2 & -6 & 1 \ -3 & 3 & -2 \end{pmatrix} \]
We solve for \( x, y, \text{and} z \) by finding determinants of modified versions of this matrix. For example, when finding \( \text{det}(A_x) \), we replace the first column of \( A \) with the constants:
\[ A_x = \begin{pmatrix} -2 & 3 & -1 \ -5 & -6 & 1 \ 5 & 3 & -2 \end{pmatrix} \]

linear algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It's a foundational subject in mathematics as well as in various applied fields like physics, engineering, and computer science.

Cramer's Rule, used in the provided exercise, is an application of linear algebra to solve systems of linear equations. It leverages the properties of determinants to find solutions. The rule states that each variable can be found using the formula:
\[ x_i = \frac{\text{det}(A_i)}{\text{det}(A)} \]
Where \( A_i \) is a matrix formed by replacing the \( i \)-th column of \( A \) with the constants from the equations. Calculating these determinants and applying the formula, we can find the solutions to the system.

In summary, by understanding these core concepts—system of equations, determinants, matrix algebra, and linear algebra—you are well-equipped to approach and solve linear systems using Cramer's Rule.