Problem 12 - Precalculus

Question

Use Cramer’s Rule to solve the system of linear equations. $$ \left\\{\begin{aligned} \frac{1}{2} x-\frac{1}{5} y &=1 \\ 6 x+7 y &=3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{76}{47}\), \(y = -\frac{45}{47}\).

Step 1: Write the system in matrix form

The system of equations can be expressed in matrix form as: \[\begin{bmatrix}\frac{1}{2} & -\frac{1}{5} \6 & 7\end{bmatrix}\begin{bmatrix}x \y\end{bmatrix}=\begin{bmatrix}1 \3\end{bmatrix}\] This is equivalent to \(A\mathbf{x} = \mathbf{b}\) where \(A\) is the coefficient matrix, \(\mathbf{x}\) is the variable vector, and \(\mathbf{b}\) is the constant vector.

Step 2: Identify the coefficient matrix

The coefficient matrix \(A\) is given by: \[A = \begin{bmatrix} \frac{1}{2} & -\frac{1}{5} \ 6 & 7 \end{bmatrix}\] This matrix represents the coefficients of \(x\) and \(y\) in the system of equations.

Step 3: Compute the determinant of matrix A

The determinant of \(A\), denoted as \(\det(A)\), is given by: \[\det(A) = \begin{vmatrix} \frac{1}{2} & -\frac{1}{5} \ 6 & 7 \end{vmatrix} = \left(\frac{1}{2} \times 7\right) - \left(-\frac{1}{5} \times 6\right)\]Calculating this, we get:\[\det(A) = \frac{7}{2} + \frac{6}{5} = \frac{35}{10} + \frac{12}{10} = \frac{47}{10}\]

Step 4: Form the matrix A_x and calculate its determinant

To find \(A_x\), replace the first column of \(A\) with \(\mathbf{b}\):\[A_x = \begin{bmatrix} 1 & -\frac{1}{5} \ 3 & 7 \end{bmatrix}\]Now, calculate \(\det(A_x)\):\[\det(A_x) = \begin{vmatrix} 1 & -\frac{1}{5} \ 3 & 7 \end{vmatrix} = (1 \times 7) - (-\frac{1}{5} \times 3) = 7 + \frac{3}{5} = \frac{35}{5} + \frac{3}{5} = \frac{38}{5}\]

Step 5: Form the matrix A_y and calculate its determinant

To find \(A_y\), replace the second column of \(A\) with \(\mathbf{b}\):\[A_y = \begin{bmatrix} \frac{1}{2} & 1 \ 6 & 3 \end{bmatrix}\]Now, calculate \(\det(A_y)\):\[\det(A_y) = \begin{vmatrix} \frac{1}{2} & 1 \ 6 & 3 \end{vmatrix} = \left(\frac{1}{2} \times 3\right) - (1 \times 6) = \frac{3}{2} - 6 = \frac{3}{2} - \frac{12}{2} = -\frac{9}{2}\]

Step 6: Apply Cramer’s Rule to find x and y

Cramer's Rule gives the solutions to the system of equations as: \[x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}\]Substituting the determinants, we have:\[x = \frac{\frac{38}{5}}{\frac{47}{10}} = \frac{38}{5} \times \frac{10}{47} = \frac{76}{47}\]\[y = \frac{-\frac{9}{2}}{\frac{47}{10}} = -\frac{9}{2} \times \frac{10}{47} = -\frac{45}{47}\]

Step 7: Conclusion

The solution to the system of equations is \(x = \frac{76}{47}\), and \(y = -\frac{45}{47}\).

Key Concepts

Determinant

The determinant is a special number that can be calculated from a square matrix. This number provides useful information about the matrix, such as whether it has an inverse, and it is pivotal in various applications, including solving systems of linear equations using Cramer's Rule. To compute the determinant of a 2x2 matrix, like the one in this exercise, you use the formula:

• For a matrix \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\], the determinant is calculated as \( ad - bc \).

In our example, computing the determinant of the coefficient matrix \[A = \begin{bmatrix} \frac{1}{2} & -\frac{1}{5} \ 6 & 7 \end{bmatrix}\], we get:\[\det(A) = \left(\frac{1}{2} \cdot 7\right) - \left(-\frac{1}{5} \cdot 6\right) = \frac{7}{2} + \frac{6}{5} = \frac{47}{10}\]This number is crucial as it indicates whether the system of equations has a unique solution. If the determinant were zero, it would mean the system might have no solution or infinitely many solutions.

System of Linear Equations

A system of linear equations consists of equations where each represents a straight line. In this context, solving the system means finding a common solution that satisfies all equations. For example, in a system like:

• \[ \begin{aligned} \frac{1}{2}x - \frac{1}{5}y &= 1 \ 6x + 7y &= 3 \end{aligned} \]

Each equation can be visualized as a line in a two-dimensional plane, and we seek the intersection point of these lines.
In solving systems of linear equations, various methods exist, including substitution, elimination, and matrix techniques such as Cramer’s Rule. The choice of method often depends on the structure and complexity of the system. Cramer's Rule is particularly useful for systems where the number of equations equals the number of unknowns and the determinant of the coefficient matrix is non-zero.

Matrix Form

Representing a system of linear equations in matrix form can simplify the process of finding solutions, especially for larger systems. In matrix form, we write the system as a single equation \(A\mathbf{x} = \mathbf{b}\), where:

• \(A\) is the coefficient matrix containing all coefficients of the variables,
• \(\mathbf{x}\) is the variable matrix (or vector) containing the unknowns, and
• \(\mathbf{b}\) is the constant matrix (or vector) with the constants from the right side of the equations.

For the given system, the matrix form is:\[\begin{bmatrix} \frac{1}{2} & -\frac{1}{5} \ 6 & 7 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 1 \ 3 \end{bmatrix}\]This representation allows us to apply matrix algebra to find the solutions more efficiently through techniques like Cramer's Rule, which relies on calculating determinants.

Solution to Equations

Finding the solution to a given system of equations involves determining the values of the variables that satisfy all equations simultaneously. Using Cramer's Rule, a direct method involving determinants, we can find the solutions for each variable when the system meets certain criteria, such as having a non-zero determinant.
Here's how Cramer's Rule applies:

• Calculate the determinant of the coefficient matrix \(\det(A)\).
• Formulate matrices like \(A_x\) and \(A_y\) by replacing columns of \(A\) with \(\mathbf{b}\), the constant matrix, and find their determinants \(\det(A_x)\) and \(\det(A_y)\).
• The solutions for the variables are given by: \[x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}\]

In this exercise, we found:

• \( x = \frac{76}{47} \)
• \( y = -\frac{45}{47} \)

These values provide the coordinates of the intersection point of the lines represented by the original equations, marking where both equations are true simultaneously.