Problem 34
Question
For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{array}{l} 8 x-2 y=-3 \\ -4 x+6 y=4 \end{array} $$
Step-by-Step Solution
Verified Answer
The solution is \(x = -\frac{1}{4}\) and \(y = \frac{1}{2}\).
1Step 1: Identify Coefficient Matrix
The given system of equations is: \\(8x - 2y = -3\) \\(-4x + 6y = 4\). \The coefficient matrix \(A\) is formed by the coefficients of \(x\) and \(y\): \\[ A = \begin{bmatrix} 8 & -2 \ -4 & 6 \end{bmatrix} \]
2Step 2: Find Determinant of Coefficient Matrix
Calculate the determinant of matrix \(A\), denoted as \( \det(A) \). Use the formula for a 2x2 matrix: \\(\text{det}(A) = ad - bc\). \Substituting the values, we get: \\(\det(A) = 8 \cdot 6 - (-2) \cdot (-4) = 48 - 8 = 40\).
3Step 3: Formulate Matrices for Cramer’s Rule
Formulate matrices \(A_x\) and \(A_y\) by replacing the respective column in \(A\) with the constants from the right-hand side of the equations. \[ A_x = \begin{bmatrix} -3 & -2 \ 4 & 6 \end{bmatrix} \] \\[ A_y = \begin{bmatrix} 8 & -3 \ -4 & 4 \end{bmatrix} \]
4Step 4: Determine \( \det(A_x) \) and \( \det(A_y) \)
Calculate \( \det(A_x) \): \\(\text{det}(A_x) = (-3)(6) - (-2)(4) = -18 + 8 = -10\). \Calculate \( \det(A_y) \): \\(\text{det}(A_y) = (8)(4) - (-3)(-4) = 32 - 12 = 20\).
5Step 5: Solve for x and y Using Cramer's Rule
Cramer's Rule states:\\( x = \frac{\det(A_x)}{\det(A)} \), \( y = \frac{\det(A_y)}{\det(A)} \). \Substitute the determinants into Cramer's formulas: \\( x = \frac{-10}{40} = -\frac{1}{4} \). \\( y = \frac{20}{40} = \frac{1}{2} \).
Key Concepts
System of Linear EquationsDeterminantCoefficient MatrixSolving Equations
System of Linear Equations
A system of linear equations involves multiple linear equations with the same set of unknowns. In the equation provided, there are two equations: \(8x - 2y = -3\) and \(-4x + 6y = 4\). Here, both equations work together to find the values of \(x\) and \(y\).
To visualize, imagine each equation represents a line on a graph. The point where the lines intersect is the solution to the system. It gives the precise values of \(x\) and \(y\) that satisfy both equations simultaneously.
Solving such systems is crucial in various fields such as physics, engineering, and economics, where you need to find variables that satisfy different conditions.
To visualize, imagine each equation represents a line on a graph. The point where the lines intersect is the solution to the system. It gives the precise values of \(x\) and \(y\) that satisfy both equations simultaneously.
Solving such systems is crucial in various fields such as physics, engineering, and economics, where you need to find variables that satisfy different conditions.
Determinant
The determinant is a special number that we can calculate from a square matrix. It provides important properties about the matrix, like whether it is invertible. In this exercise, the determinant tells us whether the linear equations have an intersecting solution. If the determinant is zero, the system does not have a unique solution.
For a 2x2 matrix like \(A=\begin{bmatrix} 8 & -2 \ -4 & 6 \end{bmatrix}\), the determinant is calculated as \(ad - bc\). Here, \(a=8\), \(b=-2\), \(c=-4\), and \(d=6\), so \(\text{det}(A) = 8 \times 6 - (-2) \times (-4) = 48 - 8 = 40\).
Since the determinant is not zero, we can apply methods like Cramer's Rule to find a unique solution to our system.
For a 2x2 matrix like \(A=\begin{bmatrix} 8 & -2 \ -4 & 6 \end{bmatrix}\), the determinant is calculated as \(ad - bc\). Here, \(a=8\), \(b=-2\), \(c=-4\), and \(d=6\), so \(\text{det}(A) = 8 \times 6 - (-2) \times (-4) = 48 - 8 = 40\).
Since the determinant is not zero, we can apply methods like Cramer's Rule to find a unique solution to our system.
Coefficient Matrix
The coefficient matrix is a matrix formed only by the coefficients of the variables in a system of equations. In our problem, the coefficient matrix \(A\) is made from the coefficients of \(x\) and \(y\) in the equations: \(8x - 2y = -3\) and \(-4x + 6y = 4\).
It is represented as: \[ A = \begin{bmatrix} 8 & -2 \ -4 & 6 \end{bmatrix} \]
This matrix is essential since it forms the basis for calculating determinants and applying Cramer's Rule. The arrangement of coefficients in structured matrices helps in algorithmically solving the system of equations efficiently.
It is represented as: \[ A = \begin{bmatrix} 8 & -2 \ -4 & 6 \end{bmatrix} \]
This matrix is essential since it forms the basis for calculating determinants and applying Cramer's Rule. The arrangement of coefficients in structured matrices helps in algorithmically solving the system of equations efficiently.
Solving Equations
Solving equations means finding the values of variables that make the equation true. With systems of linear equations, like in this exercise, the process often involves finding the intersection of multiple lines.
Cramer's Rule is a straightforward method for this, given the determinant of the coefficient matrix is non-zero. It uses determinants of matrices formed by replacing columns with solution vectors. For example, to solve \(x\) from the equation system: \[ x = \frac{\det(A_x)}{\det(A)} \] Calculating, \(x = \frac{-10}{40} = -\frac{1}{4}\).
Similarly, to solve for \(y\): \[ y = \frac{\det(A_y)}{\det(A)} \] Calculating, \(y = \frac{20}{40} = \frac{1}{2}\). Thus, the solutions are \(x = -\frac{1}{4}\) and \(y = \frac{1}{2}\), showing the intersection point for these lines.
Cramer's Rule is a straightforward method for this, given the determinant of the coefficient matrix is non-zero. It uses determinants of matrices formed by replacing columns with solution vectors. For example, to solve \(x\) from the equation system: \[ x = \frac{\det(A_x)}{\det(A)} \] Calculating, \(x = \frac{-10}{40} = -\frac{1}{4}\).
Similarly, to solve for \(y\): \[ y = \frac{\det(A_y)}{\det(A)} \] Calculating, \(y = \frac{20}{40} = \frac{1}{2}\). Thus, the solutions are \(x = -\frac{1}{4}\) and \(y = \frac{1}{2}\), showing the intersection point for these lines.
Other exercises in this chapter
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