Question

Solve each system of equations using Cramer's Rule if is applicable. If Cramer's Rule is not applicable, write, "Not applicable" \(\left\\{\begin{array}{l}x+y=8 \\ x-y=4\end{array}\right.\)

Short answer

The solution is \(x = 6\) and \(y = 2\).

Step-by-step solution

Identify Coefficient Matrix

Write the coefficient matrix for the system of equations. The coefficient matrix is formed by the coefficients of the variables x and y from both equations. \[\begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix}\]

Determine the Determinant of the Coefficient Matrix

Calculate the determinant of the coefficient matrix. \[ \text{Det}(A) = \begin{vmatrix} 1 & 1 \ 1 & -1 \end{vmatrix} = (1 \times -1) - (1 \times 1) = -1 - 1 = -2 \]

Check Cramer's Rule Applicability

Cramer's Rule is applicable if the determinant is not zero. Since \( \text{Det}(A) = -2 eq 0 \), Cramer's Rule is applicable.

Create the Matrices for Solutions

Form two new matrices by replacing one column at a time with the constants from the right-hand side of the system equations. For finding \(x\): \[ A_x = \begin{bmatrix} 8 & 1 \ 4 & -1 \end{bmatrix} \] For finding \(y\): \[ A_y = \begin{bmatrix} 1 & 8 \ 1 & 4 \end{bmatrix} \]

Calculate Determinants for Solutions

Calculate the determinants of the newly formed matrices. \[ \text{Det}(A_x) = \begin{vmatrix} 8 & 1 \ 4 & -1 \end{vmatrix} = (8 \times -1) - (1 \times 4) = -8 - 4 = -12 \] \[ \text{Det}(A_y) = \begin{vmatrix} 1 & 8 \ 1 & 4 \end{vmatrix} = (1 \times 4) - (8 \times 1) = 4 - 8 = -4 \]

Apply Cramer's Rule

Use Cramer's Rule to find the values of \(x\) and \(y\). \[ x = \frac{\text{Det}(A_x)}{\text{Det}(A)} = \frac{-12}{-2} = 6 \] \[ y = \frac{\text{Det}(A_y)}{\text{Det}(A)} = \frac{-4}{-2} = 2 \]

Verify the Solution

Substitute \(x = 6\) and \(y = 2\) back into the original equations to verify: The first equation: \(6 + 2 = 8\) The second equation: \(6 - 2 = 4\) Both equations are satisfied, so the solution is correct.

Key concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. In this exercise, we have two equations with two variables, x and y:

• x + y = 8
• x - y = 4

The goal is to find values of x and y that satisfy both equations simultaneously. When solving systems of equations, there are several methods available including substitution, elimination, and Cramer's Rule, which uses determinants.

Determinant

The determinant is a special number that can be calculated from a square matrix. It provides important information about the matrix, including whether it is invertible. For a 2x2 matrix:\[ \text{Det}(A) = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc\]In this exercise, the determinant of the coefficient matrix is calculated to check if Cramer's Rule can be applied. The determinant is found to be \-2\, which is not zero. Thus, Cramer's Rule is applicable, allowing us to solve for the variables using this method.

Coefficient Matrix

The coefficient matrix is composed of the coefficients of the variables in the system of equations. For our given equations, the coefficients of x and y are arranged into a matrix:\[ \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \]This matrix is crucial because we use its determinant to determine the applicability of Cramer's Rule. If the determinant is zero, the system either has no solution or an infinite number of solutions.

Linear Algebra

Linear algebra is a branch of mathematics concerned with vector spaces and linear mappings between them. It includes the study of systems of linear equations, matrices, determinants, vector spaces, and more. In this problem, linear algebra concepts are applied to solve the system of equations using Cramer's Rule. This involves steps such as forming the coefficient matrix, calculating its determinant, and forming new matrices to solve for each variable. Understanding these processes is essential for effectively solving systems of equations in linear algebra.