Problem 64 - A Graphical Approach to Precalculus with Limits

Question

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}x+9 y &=-15 \\\3 x+2 y &=5\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 3\) and \(y = -2\).

Step 1: Write the system in matrix form

Identify the coefficient matrix \(A\) and the constant matrices \(B_1\) and \(B_2\). The system can be written as: \[\begin{aligned} A &= \begin{bmatrix} 1 & 9 \ 3 & 2 \end{bmatrix}, \ b &= \begin{bmatrix} -15 \ 5 \end{bmatrix}. \end{aligned}\]

Step 2: Calculate the Determinant of Coefficient Matrix (D)

Calculate the determinant of matrix \(A\). \(D = \begin{vmatrix} 1 & 9 \ 3 & 2 \end{vmatrix} = 1 \times 2 - 9 \times 3 = 2 - 27 = -25.\)

Step 3: Find Determinant of Matrix A with x-replacement (Dx)

Replace the first column of \(A\) with the constants from matrix \(b\) and calculate the determinant \(D_x\). \[D_x = \begin{vmatrix} -15 & 9 \ 5 & 2 \end{vmatrix} = (-15 \times 2) - (9 \times 5) = -30 - 45 = -75.\]

Step 4: Find Determinant of Matrix A with y-replacement (Dy)

Replace the second column of \(A\) with the constants from matrix \(b\) and calculate the determinant \(D_y\). \[D_y = \begin{vmatrix} 1 & -15 \ 3 & 5 \end{vmatrix} = (1 \times 5) - (3 \times -15) = 5 + 45 = 50.\]

Step 5: Use Cramer's Rule to find x and y

Apply Cramer's Rule: \(x = \frac{D_x}{D}\) and \(y = \frac{D_y}{D}\). Substitute the values:\[x = \frac{-75}{-25} = 3,\quad y = \frac{50}{-25} = -2.\]

Step 6: Solution Confirmation

Verify the solution by substituting \(x = 3\) and \(y = -2\) into the original equations to ensure they are satisfied.

Key Concepts

System of Equations

A system of equations consists of multiple equations that are solved together because they share common variables. In the given problem, we have two equations sharing the variables \(x\) and \(y\). These are:

• \(x + 9y = -15\)
• \(3x + 2y = 5\)

The goal is to find the values of these variables that satisfy all equations at once. The unique solution of this system can be effectively found using Cramer's Rule, provided the determinant of the coefficient matrix is non-zero. If the determinant is zero, it implies either infinite solutions or no solution, and an alternative method would be required.

Determinant Calculation

The determinant is a unique number associated with square matrices and plays a crucial role in solving systems of equations. To calculate the determinant of a 2x2 matrix, apply the formula: \[ D = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc \] In our exercise, the coefficient matrix \(A\) is \(\begin{bmatrix} 1 & 9 \ 3 & 2 \end{bmatrix}\). Calculating its determinant gives: \[ D = 1 \times 2 - 9 \times 3 = 2 - 27 = -25 \] Since \(D eq 0\), Cramer's Rule is applicable to find the solution. This calculation is crucial as a zero determinant would lead us to seek other methods due to the lack of a unique solution.

Matrix Form

Representing the system of equations in matrix form is a critical step when using Cramer's Rule and makes calculations straightforward. The coefficient matrix holds the coefficients of the variables, while the constant matrix holds the constants of the system:

• Coefficient Matrix: \(A = \begin{bmatrix} 1 & 9 \ 3 & 2 \end{bmatrix}\)
• Constant Matrix: \(b = \begin{bmatrix} -15 \ 5 \end{bmatrix}\)

In this matrix representation, each row forms a linear equation, and the matrices encode all the vital information for solving the system. This form is both elegant and functional, laying the groundwork for determinant calculations.

Solution Verification

After calculating the values of \(x\) and \(y\) with Cramer's Rule, it's important to verify the solution to ensure accuracy. In the exercise, the solution \(x = 3\) and \(y = -2\) was obtained. To confirm, substitute these values back into the original equations:

• First equation: \(3 + 9(-2) = 3 - 18 = -15\)
• Second equation: \(3(3) + 2(-2) = 9 - 4 = 5\)

Both equations validate the solution, showing that our calculated values are indeed correct. Verification is a crucial final step that reinforces confidence in the obtained answers, ensuring the problem is approached with precision.