Problem 29
Question
For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{array}{l} 4 x+3 y=23 \\ 2 x-y=-1 \end{array} $$
Step-by-Step Solution
Verified Answer
The solutions are \(x = 2\) and \(y = 5\).
1Step 1: Write the System in Matrix Form
First, represent the system of equations using matrix notation: \(Ax = B\), where \(A\) is the coefficient matrix, \(x\) is the variable matrix, and \(B\) is the constant matrix.\[ A = \begin{bmatrix} 4 & 3 \ 2 & -1 \end{bmatrix}, \quad x = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 23 \ -1 \end{bmatrix} \]
2Step 2: Calculate the Determinant of Matrix A
Find the determinant of matrix \(A\), which is necessary for applying Cramer's Rule.The determinant is given by:\[ \text{det}(A) = (4)(-1) - (3)(2) = -4 - 6 = -10 \]
3Step 3: Form Matrix A_x and Calculate Its Determinant
Replace the first column of \(A\) with \(B\) to form matrix \(A_x\) and calculate its determinant.\[ A_x = \begin{bmatrix} 23 & 3 \ -1 & -1 \end{bmatrix} \]Calculate the determinant:\[ \text{det}(A_x) = (23)(-1) - (3)(-1) = -23 + 3 = -20 \]
4Step 4: Form Matrix A_y and Calculate Its Determinant
Replace the second column of \(A\) with \(B\) to form matrix \(A_y\) and calculate its determinant.\[ A_y = \begin{bmatrix} 4 & 23 \ 2 & -1 \end{bmatrix} \]Calculate the determinant:\[ \text{det}(A_y) = (4)(-1) - (23)(2) = -4 - 46 = -50 \]
5Step 5: Apply Cramer's Rule to Find x and y
Use Cramer's Rule to solve for \(x\) and \(y\). According to Cramer's Rule:\[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-20}{-10} = 2 \]\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-50}{-10} = 5 \]
6Step 6: Verify Solutions
Verify the solutions \(x = 2\) and \(y = 5\) by substituting back into the original equations:1. For \(4x + 3y = 23\): \[4(2) + 3(5) = 8 + 15 = 23\] So the equation holds.2. For \(2x - y = -1\): \[2(2) - 5 = 4 - 5 = -1\] This equation also holds, confirming our solutions.
Key Concepts
System of Linear EquationsDeterminant of a MatrixMatrix NotationSolving Equations with Matrices
System of Linear Equations
A system of linear equations is a collection of two or more linear equations with the same set of variables. These equations work together and share solutions. In our example, we have the equations:
Solving these systems can tell us the specific values of the variables – in this case, \(x\) and \(y\) – that make both equations true simultaneously.
There are several methods to solve such systems, with Cramer's Rule being particularly useful when you have the same number of equations and variables.
- \(4x + 3y = 23\)
- \(2x - y = -1\)
Solving these systems can tell us the specific values of the variables – in this case, \(x\) and \(y\) – that make both equations true simultaneously.
There are several methods to solve such systems, with Cramer's Rule being particularly useful when you have the same number of equations and variables.
Determinant of a Matrix
The determinant is a special number calculated from a square matrix. It provides important properties of the matrix and is especially useful in various mathematical applications such as solving systems of linear equations.
To find the determinant of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the formula is given by \(ad - bc\).
In our exercise, the determinant of matrix \(A\):
This determinant is crucial, as it shows us if the system has a unique solution, is inconsistent, or if the equations are dependent.
To find the determinant of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the formula is given by \(ad - bc\).
In our exercise, the determinant of matrix \(A\):
- \(A = \begin{bmatrix} 4 & 3 \ 2 & -1 \end{bmatrix}\)
This determinant is crucial, as it shows us if the system has a unique solution, is inconsistent, or if the equations are dependent.
Matrix Notation
Matrix notation is a compact way of representing a set of equations using matrices, which are rectangular arrays of numbers, symbols, or expressions arranged in rows and columns.
In our exercise, the system of equations was written in matrix form as \(Ax = B\), where:
In our exercise, the system of equations was written in matrix form as \(Ax = B\), where:
- \(A\) is the coefficient matrix \(\begin{bmatrix} 4 & 3 \ 2 & -1 \end{bmatrix}\)
- \(x\) is the variable matrix \(\begin{bmatrix} x \ y \end{bmatrix}\)
- \(B\) is the constant matrix \(\begin{bmatrix} 23 \ -1 \end{bmatrix}\)
Solving Equations with Matrices
Solving equations with matrices allows for a structured approach to find unknown variables in a system of equations. Cramer's Rule is one method specifically for uses where the system is determined – meaning the number of equations equals the number of variables.
The method involves:
The method involves:
- Calculating the determinant of the coefficient matrix \(A\).
- Forming new matrices (e.g., \(A_x\) and \(A_y\)) by replacing columns in \(A\) with the constant matrix \(B\).
- Computing determinants of these new matrices \(A_x\) and \(A_y\).
- Using these determinants to find variables with the formula \(x = \frac{\text{det}(A_x)}{\text{det}(A)}\) and \(y = \frac{\text{det}(A_y)}{\text{det}(A)}\).
Other exercises in this chapter
Problem 28
For the following exercises, use any method to solve the nonlinear system. $$ \begin{array}{l} -x^{2}+y=2 \\ -4 x+y=-1 \end{array} $$
View solution Problem 28
For the following exercises, solve each system by Gaussian elimination. $$ \begin{array}{c} 3 x-\frac{1}{2} y-z=-\frac{1}{2} \\ 4 x+z=3 \\ -x+\frac{3}{2} y=\fra
View solution Problem 29
For the following exercises, solve the system using the inverse of a \(2 \times 2\) matrix. $$ \begin{array}{l} 3 x-2 y=6 \\ -x+5 y=-2 \end{array} $$
View solution Problem 29
For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{l} 2 x-y=2 \\ 3 x+2 y=17 \end{array} $$
View solution