Chapter 7: Problem 60
Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}&4 x-y=0\\\&2 x+3 y=14\end{aligned}$$
Short Answer
Expert verified
The solution is \(x = 1\) and \(y = 4\).
Step by step solution
01
Write the System in Matrix Form
The system of equations can be written in matrix form as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables, and \(B\) is the column matrix of constants. Here, \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\), \(X = \begin{bmatrix} x \ y \end{bmatrix}\), and \(B = \begin{bmatrix} 0 \ 14 \end{bmatrix}\).
02
Calculate the Determinant of the Coefficient Matrix
The determinant of matrix \(A\), denoted as \(|A|\), is calculated as \(|A| = (4)(3) - (2)(-1) = 12 + 2 = 14\). Since \(|A| eq 0\), we can proceed with Cramer's Rule.
03
Find the Determinant of Matrix for x
To find \(x\) using Cramer's Rule, replace the first column of \(A\) with \(B\) to form a new matrix \(A_x = \begin{bmatrix} 0 & -1 \ 14 & 3 \end{bmatrix}\). Calculate the determinant of \(A_x\): \(|A_x| = (0)(3) - (14)(-1) = 0 + 14 = 14\).
04
Find the Determinant of Matrix for y
To find \(y\), replace the second column of \(A\) with \(B\) to form a new matrix \(A_y = \begin{bmatrix} 4 & 0 \ 2 & 14 \end{bmatrix}\). Calculate the determinant of \(A_y\): \(|A_y| = (4)(14) - (2)(0) = 56\).
05
Solve for x and y Using Cramer's Rule
Using Cramer's Rule, solve for \(x\) and \(y\): \(x = \frac{|A_x|}{|A|} = \frac{14}{14} = 1\) and \(y = \frac{|A_y|}{|A|} = \frac{56}{14} = 4\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
System of Equations
A system of equations is a collection of two or more equations that share a set of variables. In the given problem, we have two equations: \(4x - y = 0\) and \(2x + 3y = 14\). Each equation describes a line on a coordinate plane, and solving the system means finding the point(s) where these lines intersect, representing the solution for these variables.
There are several methods to solve systems of equations:
There are several methods to solve systems of equations:
- Substitution
- Elimination
- Matrix methods, such as Cramer's Rule and Gaussian Elimination
Determinant of a Matrix
The determinant of a matrix is a special number that you can calculate from its elements. It provides crucial information about the matrix, such as whether it is invertible. For a 2x2 matrix, which is what we are using in this example, the determinant can be calculated as follows: given the matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant of \(A\), denoted \(|A|\), is \(|A| = ad - bc\).
In our exercise, we calculate the determinant of the coefficient matrix \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\). The determinant is \( (4 \times 3) - (2 \times -1) = 12 + 2 = 14 \). Since \(|A| eq 0\), we can proceed with Cramer's Rule, signifying that the system has a unique solution.
In our exercise, we calculate the determinant of the coefficient matrix \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\). The determinant is \( (4 \times 3) - (2 \times -1) = 12 + 2 = 14 \). Since \(|A| eq 0\), we can proceed with Cramer's Rule, signifying that the system has a unique solution.
Matrix Form
To solve a system using matrices, we write it in a special format called matrix form. In matrix form, the system \(4x - y = 0\) and \(2x + 3y = 14\) can be expressed using matrices as \(AX = B\), where:
- \(A = \begin{bmatrix} 4 & -1 \ 2 & 3 \end{bmatrix}\) is the coefficient matrix representing the coefficients of the variables \(x\) and \(y\).
- \(X = \begin{bmatrix} x \ y \end{bmatrix}\) is the column matrix of variables.
- \(B = \begin{bmatrix} 0 \ 14 \end{bmatrix}\) is the column matrix of constants from the right-hand side of the equations.
Solving Equations
With Cramer's Rule, solving the system of equations involves replacing the columns of the coefficient matrix with the constant matrix one at a time to form new matrices. Each new matrix helps us find the value of one variable.
Here's how it works:
- \(x = \frac{|A_x|}{|A|} = \frac{14}{14} = 1\)
- \(y = \frac{|A_y|}{|A|} = \frac{56}{14} = 4\)
Thus, the solution \(x = 1, y = 4\) is the point where the two lines represented by the equations intersect.
Here's how it works:
- For \(x\), replace the first column of \(A\) with \(B\) to form \(A_x = \begin{bmatrix} 0 & -1 \ 14 & 3 \end{bmatrix}\). The determinant is \(|A_x| = 14\).
- For \(y\), replace the second column of \(A\) with \(B\) to form \(A_y = \begin{bmatrix} 4 & 0 \ 2 & 14 \end{bmatrix}\). The determinant is \(|A_y| = 56\).
- \(x = \frac{|A_x|}{|A|} = \frac{14}{14} = 1\)
- \(y = \frac{|A_y|}{|A|} = \frac{56}{14} = 4\)
Thus, the solution \(x = 1, y = 4\) is the point where the two lines represented by the equations intersect.