Problem 20
Question
Use Cramer's rule to solve each system or to determine that the system is inconsistent or contains dependent equations. $$ \begin{aligned}&3 x=7 y+1\\\&2 x=3 y-1\end{aligned} $$
Step-by-Step Solution
Verified Answer
The solution to the system of equations using Cramer's rule is \(x = -2\) and \(y = -1\).
1Step 1: Write the System in Standard Form
Arranging the system in standard form (Ax+By = C): \[\begin{align*}3x - 7y &= 1 \2x - 3y &= -1\end{align*}\]
2Step 2: Write Coefficient Matrix and Calculate Determinant
The coefficient matrix D and its determinant \(\Delta\) are:\[D=\begin{bmatrix} 3 & -7 \ 2 & -3 \end{bmatrix}\]To compute the determinant: \[\Delta = (3*-3) - (2*-7) = -9 - (-14) = 5\]Since \(\Delta\) is not equal to 0, this system has a unique solution.
3Step 3: Find \(x\) Using Cramer's Rule
Replace first column in D with the constants:\[D_x=\begin{bmatrix} 1 & -7 \ -1 & -3 \end{bmatrix}\]Compute determinant \(\Delta_x = (1*-3) - (-1*-7) = -3 - 7 = -10\)Thus, \(x=\Delta x/\Delta = -10/5 = -2\)
4Step 4: Find \(y\) Using Cramer's Rule
Replace second column in D with the constants:\[D_y=\begin{bmatrix} 3 & 1 \ 2 & -1 \end{bmatrix}\]Compute determinant \(\Delta_y = (3*-1) - (2*1) = -3 - 2 = -5\)Thus, \(y=\Delta y/\Delta = -5/5 = -1\)
Key Concepts
System of Linear EquationsDeterminantsAlgebraic Solutions
System of Linear Equations
Understanding the basics of a system of linear equations is essential when diving into advanced algebra. A system of linear equations consists of two or more linear equations that share a common set of variables. The goal is to find a solution that satisfies all equations in the system, which typically means finding the values of the variables that make all equations true.
In the context of our exercise, we have a system of two equations:
Using algebraic methods such as substitution, elimination, or graphical methods is common. Yet, for systems that can be complicated or not easily simplified, alternative methods like Cramer's rule – ideal for systems with a unique solution – can be very efficient.
In the context of our exercise, we have a system of two equations:
- 3x - 7y = 1
- 2x - 3y = -1
Using algebraic methods such as substitution, elimination, or graphical methods is common. Yet, for systems that can be complicated or not easily simplified, alternative methods like Cramer's rule – ideal for systems with a unique solution – can be very efficient.
Determinants
The determinant is a special numerical value that can be calculated from the coefficients of a square matrix in linear algebra. It is often denoted as \(\Delta \) or 'det'. The significance of the determinant lies in the information it provides about the matrix. For example, a nonzero determinant implies that the matrix is invertible, and that the system of linear equations it represents has a unique solution. Conversely, a determinant of zero implies that the matrix is not invertible, and the system may have no solution or an infinite number of solutions.
In our exercise, we calculate the determinant of the coefficient matrix to use Cramer's rule. If we have a 2x2 matrix, the determinant is straightforward to compute based on a simple formula:\[\begin{equation}\[\Delta = a_{11} a_{22} - a_{12} a_{21}\]\end{equation}\]This single value plays a crucial role in determining the existence and uniqueness of solutions to the system of equations. As we can see from our example, the determinant \( \Delta = 5 \) is non-zero, indicating a unique solution exists for our system.
In our exercise, we calculate the determinant of the coefficient matrix to use Cramer's rule. If we have a 2x2 matrix, the determinant is straightforward to compute based on a simple formula:\[\begin{equation}\[\Delta = a_{11} a_{22} - a_{12} a_{21}\]\end{equation}\]This single value plays a crucial role in determining the existence and uniqueness of solutions to the system of equations. As we can see from our example, the determinant \( \Delta = 5 \) is non-zero, indicating a unique solution exists for our system.
Algebraic Solutions
Algebraic solutions involve manipulating equations using algebraic operations to find the values of unknown variables. In our textbook problem, we apply Cramer's rule, a powerful algebraic tool designed to solve systems of linear equations. It provides a direct way to calculate each variable's value without having to manipulate the equations excessively.
Cramer's rule states that in a system of linear equations with the same number of equations as unknowns, the solution can be found using determinants. Each variable is solved by replacing the respective column of the coefficient matrix with the constants from the right side of the equations, and then taking the ratio of this modified determinant to the original determinant. Following our steps:
Cramer's rule states that in a system of linear equations with the same number of equations as unknowns, the solution can be found using determinants. Each variable is solved by replacing the respective column of the coefficient matrix with the constants from the right side of the equations, and then taking the ratio of this modified determinant to the original determinant. Following our steps:
- Find the determinants \( \Delta, \Delta_x, \Delta_y \)
- Calculate the variables using \( x = \dfrac{\Delta_x}{\Delta} \) and \( y = \dfrac{\Delta_y}{\Delta} \)
Other exercises in this chapter
Problem 19
In Exercises \(19-24\), perform each matrix row operation and write the new matrix. \(\left[\begin{array}{rrr|r}2 & -6 & 4 & 10 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 &
View solution Problem 20
In Exercises \(17-26,\) let $$ A=\left[\begin{array}{rr} -3 & -7 \\ 2 & -9 \\ 5 & 0 \end{array}\right] \text { and } B=\left[\begin{array}{rr} -5 & -1 \\ 0 & 0
View solution Problem 20
find \(A^{-1}\) by forming \([A | I]\) and then using row operations to obtain [ \(I | B],\) where \(A^{-1}=[B]\) Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $
View solution Problem 20
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$\begin{aligned}w+x-y+z &=-2 \\\2 w-x+2 y-z &=7 \
View solution