Problem 59
Question
Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{array}{l}4 x+3 y=-7 \\\2 x+3 y=-11\end{array}$$
Step-by-Step Solution
Verified Answer
The solution is \(x = 2\), \(y = -5\).
1Step 1: Set up the system of equations
The given system of equations is:\[ \begin{align*} 4x + 3y &= -7 \ 2x + 3y &= -11 \end{align*} \] We will solve this using Cramer's Rule.
2Step 2: Write the coefficient matrix and calculate its determinant
The coefficient matrix \(A\) is \( \begin{bmatrix} 4 & 3 \ 2 & 3 \end{bmatrix} \). The determinant of \(A\) is calculated as follows: \[ D = \begin{vmatrix} 4 & 3 \ 2 & 3 \end{vmatrix} = (4 \times 3) - (3 \times 2) = 12 - 6 = 6 \]Since \(D eq 0\), Cramer's Rule can be applied.
3Step 3: Set up matrices to find \(x\) and \(y\)
To find \(x\), we replace the first column of \(A\) with the constants from the right side of the equations: \[A_x = \begin{bmatrix} -7 & 3 \ -11 & 3 \end{bmatrix}\] To find \(y\), we replace the second column of \(A\) with the constants: \[A_y = \begin{bmatrix} 4 & -7 \ 2 & -11 \end{bmatrix}\]
4Step 4: Calculate determinants of updated matrices
The determinant of \(A_x\) is: \[D_x = \begin{vmatrix} -7 & 3 \ -11 & 3 \end{vmatrix} = (-7 \times 3) - (3 \times -11) = -21 + 33 = 12\] The determinant of \(A_y\) is: \[D_y = \begin{vmatrix} 4 & -7 \ 2 & -11 \end{vmatrix} = (4 \times -11) - (-7 \times 2) = -44 + 14 = -30\]
5Step 5: Apply Cramer's Rule and solve for \(x\) and \(y\)
Cramer's Rule states that \(x = \frac{D_x}{D}\) and \(y = \frac{D_y}{D}\). Using our determinants: \[ x = \frac{12}{6} = 2\]\[ y = \frac{-30}{6} = -5\]Thus, the solution to the system of equations is \(x = 2\), \(y = -5\).
Key Concepts
Systems of EquationsDeterminantsCoefficient MatrixLinear Algebra
Systems of Equations
A system of equations is essentially a group of two or more equations that share common variables. In our example, we have two equations:
Systems of equations can be solved using various methods, like substitution, elimination, or, as demonstrated here, Cramer's Rule. Each method offers different insights and can be more or less convenient depending on the particular problem. Cramer's Rule is especially useful when the determinant is nonzero, allowing a direct computation of the solutions.
- The first equation: \(4x + 3y = -7\)
- The second equation: \(2x + 3y = -11\)
Systems of equations can be solved using various methods, like substitution, elimination, or, as demonstrated here, Cramer's Rule. Each method offers different insights and can be more or less convenient depending on the particular problem. Cramer's Rule is especially useful when the determinant is nonzero, allowing a direct computation of the solutions.
Determinants
Determinants play a crucial role when working with Cramer's Rule. They are numerical values calculated from square matrices and provide insightful properties.
\[ D = ad - bc\]where \(a\), \(b\), \(c\), and \(d\) are the elements of the matrix in the form \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \). These calculations are significant as they provide the means to solve for unknown variables.
- The determinant of a matrix provides a measure of how "invertible" the matrix is.
- In our example, the determinant of the coefficient matrix \(A\) is 6.
- If this determinant were zero, Cramer's Rule couldn't be applied.
\[ D = ad - bc\]where \(a\), \(b\), \(c\), and \(d\) are the elements of the matrix in the form \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \). These calculations are significant as they provide the means to solve for unknown variables.
Coefficient Matrix
The coefficient matrix, denoted by \(A\), is composed of the coefficients in front of the variables in our system of equations. For the given problem:
Such a structured approach to organizing and understanding the components of our system makes it easier to visualize and compute solutions systematically. It paves the way for linking algebraic equations to linear algebra's broader aspects.
- The coefficient matrix \(A\) is \( \begin{bmatrix} 4 & 3 \ 2 & 3 \end{bmatrix} \).
- This matrix captures the essential linear relationships articulated by the system.
Such a structured approach to organizing and understanding the components of our system makes it easier to visualize and compute solutions systematically. It paves the way for linking algebraic equations to linear algebra's broader aspects.
Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear transformations. Among its many operations, it includes systems of linear equations, matrices, determinants, and more.
One of the key features of linear algebra is employing matrices to solve systems of equations, like we did using Cramer's Rule. Here’s why this field is immensely practical:
One of the key features of linear algebra is employing matrices to solve systems of equations, like we did using Cramer's Rule. Here’s why this field is immensely practical:
- It transforms complex systems into more manageable forms.
- Through matrix operations, we can unravel solutions with increased efficiency.
- It provides tools like determinants, essential for solving and validating solutions.
Other exercises in this chapter
Problem 59
Find the equation of the parabola with vertical axis that passes through the data points shown or specified. Check your answer. $$(1.5,6.25),(0,-2),(-1.5,3.25)$
View solution Problem 59
Find each matrix product if possible. $$\left[\begin{array}{rr} 5 & 2 \\ -1 & 4 \end{array}\right]\left[\begin{array}{rr} 3 & -2 \\ 1 & 0 \end{array}\right]$$
View solution Problem 60
Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{array}{r}3 x-y=4 \\\x+y=0\end{array}$$
View solution Problem 60
Graph the solution set of each system of inequalities by hand. $$\begin{array}{r} \ln x-y \geq 1 \\ x^{2}-2 x-y \leq 1 \end{array}$$
View solution