Problem 29 - Precalculus

Question

Use Cramer's rule to solve each system of equations, if possible. $$\begin{aligned} &y=17 x+7\\\ &y=-15 x+7 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 0\) and \(y = 7\).

Step 1: Convert to Standard Form

The given system of equations needs to be in the standard form, \[ ax + by = c \].For the first equation, \( y = 17x + 7 \) can be rearranged to \[ -17x + y = 7 \]. For the second equation, \( y = -15x + 7 \), rearrange to \[ 15x + y = 7 \].Thus, the system of equations is:\[-17x + y = 7\]\[15x + y = 7\].

Step 2: Set Up the Determinant Matrix

To apply Cramer's Rule, we first need to find the determinant of the coefficient matrix. The coefficients of the variables \(x\) and \(y\) in the system form the matrix:\[A = \begin{bmatrix} -17 & 1 \ 15 & 1 \end{bmatrix}\].

Step 3: Calculate the Determinant of the Matrix A

The determinant of the matrix \(A\) is calculated as:\[\text{det}(A) = (-17)(1) - (1)(15) = -17 - 15 = -32\].

Step 4: Check if the Determinant is Zero

For Cramer's Rule to be applicable, the determinant must not be zero. Here,\[ \text{det}(A) = -32 eq 0 \]. So, Cramer's Rule can be applied.

Step 5: Calculate Determinants for x and y

Replace the columns of matrix \(A\) with the constant terms to find matrices \(D_x\) and \(D_y\):For \(x\): \[D_x = \begin{bmatrix} 7 & 1 \ 7 & 1 \end{bmatrix} \Rightarrow \text{det}(D_x) = (7)(1) - (1)(7) = 0\].For \(y\): \[D_y = \begin{bmatrix} -17 & 7 \ 15 & 7 \end{bmatrix} \Rightarrow \text{det}(D_y) = (-17)(7) - (15)(7) = -119 - 105 = -224\].

Step 6: Apply Cramer's Rule to Find x and y

Using Cramer's Rule:For \(x\): \[ x = \frac{\text{det}(D_x)}{\text{det}(A)} = \frac{0}{-32} = 0 \].For \(y\): \[ y = \frac{\text{det}(D_y)}{\text{det}(A)} = \frac{-224}{-32} = 7\].

Key Concepts

Determinant of a Matrix

The determinant is a special number calculated from a square matrix. It holds significant importance in linear algebra, particularly when dealing with systems of linear equations and transformations. To compute the determinant of a 2x2 matrix, like matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), simply use the formula: \( \text{det}(A) = ad - bc \).
For our problem, the coefficient matrix \( A = \begin{bmatrix} -17 & 1 \ 15 & 1 \end{bmatrix} \) had its determinant calculated as \( -17 \times 1 - 1 \times 15 = -32 \).
If the determinant is zero, the matrix is singular, indicating the system cannot be solved using methods like Cramer's Rule. However, if it is non-zero, like in our problem, the matrix is invertible.

A non-zero determinant signals that the linear equations involved have a unique solution, thus confirming that Cramer's Rule can be effectively applied.

System of Equations

A system of equations consists of multiple equations that must be solved together, and each equation shares certain variables. In our example, we have the system consisting of: - \( -17x + y = 7 \) - \( 15x + y = 7 \)

Each represents a line in the coordinate plane. The solution to the system represents the point where these lines intersect, if such a point exists.

Systems can have:

• one solution (lines intersect at one point),
• no solution (lines are parallel and never meet), or
• infinitely many solutions (lines overlap).

For the current exercise, using Cramer's Rule, we identified a unique solution existing as the lines intersect at the point \((0, 7)\). This reflects the conditions for applying Cramer’s Rule successfully.

Linear Equations

Linear equations form the foundation of a system of equations. These equations involve variables such as \( x \) and \( y \) in first-degree terms. Each linear equation depicts a straight line on a graph. In our solved problem, the linear equations were written in the standard form after conversion from slope-intercept form to adhere to the required format for Cramer's Rule, which is: \( ax + by = c \).

For instance, - the equation \( y = 17x + 7 \) was transformed into \( -17x + y = 7 \), - and the equation \( y = -15x + 7 \) became \( 15x + y = 7 \).

By writing the equations this way, it aligns with matrix computations essential for applying Cramer's Rule or any matrix-based solutions effectively. Keep in mind that the link between linear equations and algebra is critical, as they form the basis for understanding more complex math concepts as students advance.