Problem 43 - Precalculus: Mathematics for Calculus

Question

Use Cramer's Rule to solve the system. $$\left\\{\begin{aligned} x-6 y &=3 \\ 3 x+2 y &=1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

\( x = \frac{3}{5} \), \( y = -\frac{2}{5} \)

Step 1: Identify the Coefficients

For the given system of equations, we can identify the coefficients of the variables. The equations are: \( x - 6y = 3 \) and \( 3x + 2y = 1 \). The coefficient matrix \( A \) is \( \begin{bmatrix} 1 & -6 \ 3 & 2 \end{bmatrix} \). The constants are \( \begin{bmatrix} 3 \ 1 \end{bmatrix} \).

Step 2: Calculate the Determinant of the Coefficient Matrix

To apply Cramer's Rule, first find the determinant of matrix \( A \), denoted as \( \text{det}(A) \). \[\text{det}(A) = \begin{vmatrix} 1 & -6 \ 3 & 2 \end{vmatrix} = 1 \cdot 2 - (-6) \cdot 3 = 2 + 18 = 20\]

Step 3: Substitute and Calculate Determinant for \( x \)

To find \( x \), replace the first column of \( A \) with the constants, resulting in the matrix:\[A_x = \begin{bmatrix} 3 & -6 \ 1 & 2 \end{bmatrix}\]Calculate \( \text{det}(A_x) \):\[\text{det}(A_x) = \begin{vmatrix} 3 & -6 \ 1 & 2 \end{vmatrix} = 3 \cdot 2 - (-6) \cdot 1 = 6 + 6 = 12\]

Step 4: Substitute and Calculate Determinant for \( y \)

To find \( y \), replace the second column of \( A \) with constants:\[A_y = \begin{bmatrix} 1 & 3 \ 3 & 1 \end{bmatrix}\]Calculate \( \text{det}(A_y) \):\[\text{det}(A_y) = \begin{vmatrix} 1 & 3 \ 3 & 1 \end{vmatrix} = 1 \cdot 1 - 3 \cdot 3 = 1 - 9 = -8\]

Step 5: Solve for \( x \) and \( y \)

Cramer's Rule gives the solutions for \( x \) and \( y \) using the determinants.\[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{12}{20} = \frac{3}{5} \]\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-8}{20} = -\frac{2}{5} \]Thus, the solution to the system is \( x = \frac{3}{5} \) and \( y = -\frac{2}{5} \).

Key Concepts

Understanding Systems of Equations

When we talk about systems of equations, we're usually referring to a set of two or more equations that share common variables. These variables need to be solved in a way that satisfies all equations at the same time.

In the case of our original exercise, the system consists of two equations:

• \( x - 6y = 3 \)
• \( 3x + 2y = 1 \)

We want to find a set of values for \( x \) and \( y \) that makes both equations true simultaneously.

There are several methods to solve systems of equations, such as substitution, elimination, and graphing. However, when using Cramer's Rule, we leverage the power of matrices and determinants, making it a particularly elegant method for solving linear systems when the number of equations matches the number of unknowns.

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. This number provides important information about the matrix. In the context of solving systems of equations, the determinant can tell us whether a unique solution exists.

For a 2x2 matrix, the determinant \( \text{det}(A) \) is calculated as:

• \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \)
• The determinant is \( ad - bc \)

In our example, matrix \( A \) is:

• \( \begin{bmatrix} 1 & -6 \ 3 & 2 \end{bmatrix} \)
• Thus, \( \text{det}(A) = 1 \times 2 - (-6) \times 3 = 20 \)

The determinant is non-zero \((20)\), which means the system of equations has a unique solution.

Exploring Matrix Algebra

Matrix algebra is a powerful tool used to handle numerous mathematical computations and operations. It involves matrices, which are rectangular arrays of numbers.

Some key operations in matrix algebra include addition, subtraction, multiplication, and finding the inverse. In solving systems of equations via Cramer's Rule, we specifically use the computations of determinants, as mentioned earlier.

To apply Cramer's Rule, we structure the equation system into a coefficient matrix \( A \) and a constants vector. Here, the coefficient matrix \( A \) for our system is:

• \( \begin{bmatrix} 1 & -6 \ 3 & 2 \end{bmatrix} \)

This matrix allows us to manipulate the system mathematically, switching columns with vectors derived from the constants to calculate the individual determinants necessary for solving for each variable.

This structured approach not only offers a methodical way of finding solutions but also highlights the intersection of matrix operations and linear algebra in practical problem-solving.