Problem 35 - Algebra and Trigonometry

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: Rewrite the System in Matrix Form

Convert the system of equations into the matrix form. The system \[x - 6y = 3 \3x + 2y = 1 \]can be written as:\[A \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 3 \ 1 \end{bmatrix},\]where \[A = \begin{bmatrix} 1 & -6 \ 3 & 2 \end{bmatrix}.\]

Step 2: Calculate the Determinant of Matrix A

First, compute the determinant of matrix \(A\):\[\text{det}(A) = \begin{vmatrix} 1 & -6 \ 3 & 2 \end{vmatrix} = (1)(2) - (3)(-6) = 2 + 18 = 20.\]

Step 3: Create Matrices for Cramer's Rule

Form matrices \(A_x\) and \(A_y\) by replacing the columns of \(A\) with the constant matrix:For \(A_x\):\[A_x = \begin{bmatrix} 3 & -6 \ 1 & 2 \end{bmatrix}.\]For \(A_y\):\[A_y = \begin{bmatrix} 1 & 3 \ 3 & 1 \end{bmatrix}.\]

Step 4: Calculate Determinants for Cramer's Rule

Calculate the determinants of \(A_x\) and \(A_y\):\(\text{det}(A_x) = \begin{vmatrix} 3 & -6 \ 1 & 2 \end{vmatrix} = (3)(2) - (1)(-6) = 6 + 6 = 12.\)\(\text{det}(A_y) = \begin{vmatrix} 1 & 3 \ 3 & 1 \end{vmatrix} = (1)(1) - (3)(3) = 1 - 9 = -8.\)

Step 5: Solve for x and y Using Cramer's Rule

Using Cramer's Rule:\[x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{12}{20} = \frac{3}{5}.\]\[y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-8}{20} = -\frac{2}{5}.\]

Step 6: Conclude the Solution

The values for the variables \(x\) and \(y\) are:\(x = \frac{3}{5}\) and \(y = -\frac{2}{5}\).

Key Concepts

Matrix Form

When solving systems of equations using Cramer's Rule, the first step is to express the system in matrix form. This is essential as it transforms the system into a format that can be solved systematically using matrices.
For the given system of equations: \[ \[\[\begin{align*} x - 6y &= 3 \ 3x + 2y &= 1 \end{align*}\]\] \] We can represent it in matrix form as follows: \[ A \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 3 \ 1 \end{bmatrix} \] where the matrix \( A \) is \[ A = \begin{bmatrix} 1 & -6 \ 3 & 2 \end{bmatrix} \] Here, matrix \( A \) contains the coefficients of the variables \( x \) and \( y \), while the vector on the right side represents the constants from the equations.
Using matrix form allows us to apply various matrix operations and theorems, such as Cramer's Rule, to find the solutions efficiently.

Determinant Calculation

The determinant of a matrix is a special number that can tell us if a system of equations has a unique solution. It also plays a key role in Cramer's Rule. For our given matrix \( A \): \[ A = \begin{bmatrix} 1 & -6 \ 3 & 2 \end{bmatrix} \] We need to calculate its determinant, \( \text{det}(A) \). The formula for the determinant of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \[ \text{det}(A) = ad - bc \] Applying this to our matrix gives: \[ \text{det}(A) = (1)(2) - (3)(-6) \] \( = 2 + 18 = 20 \)
The determinant is non-zero (20), which means the system of equations has a unique solution. This computation is crucial, as a zero determinant would imply dependence among equations and possibly no unique solutions exist.

System of Equations

Understanding a system of equations involves knowing how multiple equations interact with each other to solve for unknown variables. The system \[ \[\[\begin{align*} x - 6y &= 3 \ 3x + 2y &= 1 \end{align*}\]\] \] is part of a linear system where each equation represents a line in the plane. The solution to the system is the point \((x, y)\) where these lines intersect.
Traditionally, solving a system may involve substitution or elimination, but using Cramer's Rule streamlines the process when feasible. Here, each variable is isolated through matrix manipulation, borrowing linear algebra techniques to directly find clear numerical solutions.
Recognizing this system as linear and expressing it in matrix form helps us use powerful mathematical methods and avoid more cumbersome algebraic manipulations.

Matrices Ax and Ay

To utilize Cramer's Rule, we need to form matrices \( A_x \) and \( A_y \). These matrices are variations of matrix \( A \), where columns are replaced by the constants, helping solve for each variable individually.
For matrix \( A_x \), replace the first column of \( A \) with the constants: \[ A_x = \begin{bmatrix} 3 & -6 \ 1 & 2 \end{bmatrix} \]
For matrix \( A_y \), replace the second column with the constants: \[ A_y = \begin{bmatrix} 1 & 3 \ 3 & 1 \end{bmatrix} \]
By calculating the determinants of these matrices: \[ \text{det}(A_x) = (3)(2) - (1)(-6) = 6 + 6 = 12 \] \[ \text{det}(A_y) = (1)(1) - (3)(3) = 1 - 9 = -8 \]
We can then find the solution using Cramer's Rule: \[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{12}{20} = \frac{3}{5} \] \[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-8}{20} = -\frac{2}{5} \]
This step-by-step approach showcases how separate matrices relate to specific variables in solving the system efficiently.