Question

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

Short answer

x = 3.67, y = 0.67

Step-by-step solution

- Understand the System of Equations

The given system of equations is: 1. \(x + 2y = 5\) 2. \(x - y = 3\)

- Write the Coefficient Matrix

Identify the coefficients of the variables from each equation. The coefficient matrix is: \[\begin{bmatrix} 1 & 2 \ 1 & -1 \ \end{bmatrix}\]

- Calculate the Determinant of the Coefficient Matrix (D)

Find the determinant of the coefficient matrix. Using the determinant formula for a 2x2 matrix \[\begin{vmatrix} a & b \ c & d \ \end{vmatrix} = ad - bc\]: \ D = 1(-1) - 2(1) = -1 - 2 = -3 \ \ Hence, \( D = -3 \)

- Construct the Matrices for Dx and Dy

Replace the coefficients of \(x\) and \(y\) with the constants from the equations to form new matrices: \ For Dx: \ \[\begin{bmatrix} 5 & 2 \ 3 & -1 \ \end{bmatrix}\] For Dy: \ \[\begin{bmatrix} 1 & 5 \ 1 & 3 \ \end{bmatrix}\]

- Calculate the Determinants Dx and Dy

Use the determinant formula for each matrix. For Dx: \[\begin{vmatrix} 5 & 2 \ 3 & -1 \ \end{vmatrix} = 5(-1) - 2(3) = -5 - 6 = -11 Hence, \( Dx = -11 \) \] For Dy: \[\begin{vmatrix} 1 & 5 \ 1 & 3 \ \end{vmatrix} = 1(3) - 5(1) = 3 - 5 = -2 Hence, \( Dy = -2 \) \]

- Apply Cramer's Rule to Solve for x and y

Cramer's Rule states that \( x = \frac{Dx}{D} \) and \( y = \frac{Dy}{D} \) Using the values calculated: \ x = \frac{-11}{-3} = 3.67 \ y = \frac{-2}{-3} = 0.67 \

Key concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. The goal is to find values for these variables that satisfy all the equations simultaneously. In our example, we have two equations:
1. \(x + 2y = 5\)
2. \(x - y = 3\)
These equations form a system because they share the same variables, \(x\) and \(y\). To solve them, we need to find the values of \(x\) and \(y\) that make both equations true at the same time.

Determinant

In matrix algebra, the determinant is a special number that can be calculated from a square matrix. For a 2x2 matrix, the determinant helps determine if the matrix has an inverse.
The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \ \end{bmatrix} \) is calculated as:
\[\text{det}(A) = ad - bc\]
In our exercise, the coefficient matrix is \( \begin{bmatrix} 1 & 2 \ 1 & -1 \ \end{bmatrix} \). The determinant is computed as follows:
\[D = 1*(-1) - 2*(1) = -1 - 2 = -3\]
This determinant is essential in applying methods like Cramer's Rule to solve the system of equations.

Matrix Algebra

Matrix algebra involves mathematical operations with matrices. In the context of solving linear equations, matrices simplify complex systems, making them easier to solve.
The key steps include:

• Writing the coefficient matrix from the system of equations.
• Calculating the determinant to see if a unique solution exists.
• Constructing matrices where specific columns are replaced with constants from the system for further calculations.

For our problem, the coefficient matrix is \( \begin{bmatrix} 1 & 2 \ 1 & -1 \ \end{bmatrix} \). We also form new matrices for \(Dx\) and \(Dy\) by replacing columns with constants:
- \(Dx = \begin{bmatrix} 5 & 2 \ 3 & -1 \ \end{bmatrix}\)
- \(Dy = \begin{bmatrix} 1 & 5 \ 1 & 3 \ \end{bmatrix}\)

Solving Linear Equations

To solve linear equations using Cramer's Rule, follow these steps:

• Calculate the determinant of the coefficient matrix to ensure it is non-zero.
• Create matrices for replacement variables (\(Dx\) and \(Dy\)) by replacing appropriate columns with constants from the equations.
• Find the determinants of these new matrices.
• Use the formula \(x = \frac{Dx}{D}\) and \(y = \frac{Dy}{D}\) to solve for the variables.

For our system:
\[ D = -3 \]
\[ Dx = -11 \]
\[ Dy = -2 \]
Applying Cramer's Rule:
\[ x = \frac{Dx}{D} = \frac{-11}{-3} \approx 3.67 \]
\[ y = \frac{Dy}{D} = \frac{-2}{-3} \approx 0.67 \]
This method efficiently finds the values of \(x\) and \(y\) that satisfy both original equations.