Problem 5 - Advanced Engineering Mathematics

Question

Solve the given system of equations by Cramer's rule. $$ \begin{aligned} &2 x+y=1 \\ &3 x+2 y=-2 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( x = 4 \) and \( y = -7 \).

Step 1: Write the system in matrix form

The given system of equations is \( \begin{aligned} 2x + y &= 1 \ 3x + 2y &= -2 \end{aligned} \). Rewrite this system in matrix form as \( A \mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the vector of variables, and \( \mathbf{b} \) is the constant vector. Here,\[ A = \begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 1 \ -2 \end{bmatrix}. \]

Step 2: Calculate the determinant of A

We need to find the determinant of matrix \( A \), denoted as \( \det(A) \), which is given by:\[ \det(A) = \begin{vmatrix} 2 & 1 \ 3 & 2 \end{vmatrix} = (2)(2) - (1)(3) = 4 - 3 = 1. \]

Step 3: Create and calculate determinants of modified matrices

To find \( x \), replace the first column of \( A \) with \( \mathbf{b} \), creating matrix \( A_x \):\[ A_x = \begin{bmatrix} 1 & 1 \ -2 & 2 \end{bmatrix}. \]Calculate \( \det(A_x) \):\[ \det(A_x) = \begin{vmatrix} 1 & 1 \ -2 & 2 \end{vmatrix} = (1)(2) - (1)(-2) = 2 + 2 = 4. \]To find \( y \), replace the second column of \( A \) with \( \mathbf{b} \), creating matrix \( A_y \):\[ A_y = \begin{bmatrix} 2 & 1 \ 3 & -2 \end{bmatrix}. \]Calculate \( \det(A_y) \):\[ \det(A_y) = \begin{vmatrix} 2 & 1 \ 3 & -2 \end{vmatrix} = (2)(-2) - (3)(1) = -4 - 3 = -7. \]

Step 4: Apply Cramer's rule to find the solution

By Cramer's Rule, the solution to the system is given by \( x = \frac{\det(A_x)}{\det(A)} \) and \( y = \frac{\det(A_y)}{\det(A)} \).Calculate \( x \):\[ x = \frac{\det(A_x)}{\det(A)} = \frac{4}{1} = 4. \]Calculate \( y \):\[ y = \frac{\det(A_y)}{\det(A)} = \frac{-7}{1} = -7. \]

Step 5: Conclusion: Verify the solution

The solution for the system of equations is \( x = 4 \) and \( y = -7 \). To verify, substitute back into the original equations:For \( 2x + y = 1 \):\[ 2(4) + (-7) = 8 - 7 = 1, \] which is correct.For \( 3x + 2y = -2 \):\[ 3(4) + 2(-7) = 12 - 14 = -2, \] which is also correct. Therefore, the solution satisfies the equations.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations involving the same set of variables. In our example, the system consists of two linear equations with variables \(x\) and \(y\):

• \(2x + y = 1\)
• \(3x + 2y = -2\)

The goal is to find a common pair of \((x, y)\) values that satisfy both equations simultaneously. Solving such systems can be challenging, but they are foundational in algebra, often representing intersections of lines on a graph. Different methods exist to solve these systems, with Cramer's Rule being a specific approach for systems where an equal number equations and variables are present.

Determinant Calculation

Calculating a determinant is an important step when solving systems using Cramer's Rule. A determinant is a special number calculated from a square matrix (a matrix with the same number of rows and columns). For a 2x2 matrix like \(A\): \[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix}, \] the determinant is calculated as: \[ \det(A) = ad - bc. \]In our exercise, the matrix \(A\) is: \[ \begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix}, \] and the determinant \( \det(A) \) is:

• \((2)(2) - (1)(3) = 4 - 3 = 1\)

The determinant provides insights into the matrix, such as whether it is invertible, and is crucial for applying Cramer's Rule.

Matrix Form

Expressing a system of linear equations in matrix form is a powerful technique that simplifies complex calculations. In this form, the system \( \begin{aligned} 2x + y &= 1 \ 3x + 2y &= -2 \end{aligned} \) is represented as:

• Coefficient matrix, \( A = \begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix} \)
• Variable vector, \( \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix} \)
• Constant vector, \( \mathbf{b} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \)

Thus, the system is written as \( A \mathbf{x} = \mathbf{b} \). This format allows for efficient application of linear algebra techniques like determinants and matrix operations, making complex problem-solving more structured and systematic.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, matrices, and vector spaces. It provides tools for solving systems of equations, like in our current problem using Cramer's Rule.
Key concepts include:

• Vectors: Entities representing magnitude and direction, like the variable and constant vectors in our matrix form.
• Matrices: Arrays of numbers representing coefficients of the system, operationalized via determinants for solution finding.
• Determinants: Scalar values derived from matrices that help determine solution existence and uniqueness.

By leveraging these linear algebra components, we can manipulate and solve systems like the one in the exercise, allowing us to find precise solutions through structured procedures.