Chapter 9: Problem 24
Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix, as in Example 6. Use the inverses from Exercises 7-10,15, 16,19, and 21. $$\left\\{\begin{array}{l}3 x+4 y=10 \\ 7 x+9 y=20\end{array}\right.$$
Short Answer
Expert verified
The solution is \( x = -10 \), \( y = 10 \).
Step by step solution
01
Write the System of Equations
The given system of equations is:\[\begin{align*}3x + 4y &= 10 \7x + 9y &= 20\end{align*}\]
02
Express as a Matrix Equation
We want to express the system in the form \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable matrix, and \( \mathbf{b} \) is the constants matrix:\[A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 10 \ 20 \end{bmatrix}\]So, the system becomes:\[ \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 10 \ 20 \end{bmatrix} \]
03
Find the Inverse of the Coefficient Matrix
To solve \( A\mathbf{x} = \mathbf{b} \) using the inverse matrix method, find \( A^{-1} \). The formula for the inverse of a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is:\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For our matrix \( A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \), compute the determinant:\[ \det(A) = 3 \times 9 - 4 \times 7 = 27 - 28 = -1\]And the inverse is:\[A^{-1} = \frac{1}{-1} \begin{bmatrix} 9 & -4 \ -7 & 3 \end{bmatrix} = \begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix}\]
04
Apply the Inverse Matrix to Solve for Variables
Multiply both sides of the equation \( A\mathbf{x} = \mathbf{b} \) by \( A^{-1} \):\[A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}\]This simplifies to \( \mathbf{x} = A^{-1}\mathbf{b} \) since \( A^{-1}A = I \), the identity matrix. Compute \( A^{-1}\mathbf{b} \):\[\begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix} \begin{bmatrix} 10 \ 20 \end{bmatrix} = \begin{bmatrix} (-9 \times 10) + (4 \times 20) \ (7 \times 10) + (-3 \times 20) \end{bmatrix} = \begin{bmatrix} -90 + 80 \ 70 - 60 \end{bmatrix} = \begin{bmatrix} -10 \ 10 \end{bmatrix}\]Thus, \( \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -10 \ 10 \end{bmatrix} \), meaning \( x = -10 \) and \( y = 10 \).
05
Verify the Solution
Substitute \( x = -10 \) and \( y = 10 \) back into the original equations to verify:1. \( 3(-10) + 4(10) = -30 + 40 = 10 \) (True)2. \( 7(-10) + 9(10) = -70 + 90 = 20 \) (True)The solution is verified.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Equation
To solve a system of equations using matrices, we first translate it into a form that utilizes matrix operations. This is called forming a "matrix equation." For example, consider the given system of equations:
- \(3x + 4y = 10\)
- \(7x + 9y = 20\)
Inverse of a Matrix
Solving matrix equations typically involves finding the inverse of the matrix containing the coefficients, known as the "coefficient matrix." Only square matrices (n x n) can have inverses. The inverse of a matrix \(A\), written as \(A^{-1}\), conveniently helps us isolate the variables and solve the equation.For a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse is calculated using the formula:\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]This formula involves using the determinant, \(ad - bc\), which must be non-zero to have an inverse.In our example:- Matrix \(A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix}\)- Inverse \(A^{-1} = \frac{1}{-1} \begin{bmatrix} 9 & -4 \ -7 & 3 \end{bmatrix} = \begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix}\)The inverse allows you to solve the system by multiplying both sides of the matrix equation by \(A^{-1}\).
Determinant Calculation
A determinant is a special number that can be calculated from a square matrix. It helps to understand if the matrix has an inverse, which is crucial for solving matrix equations.The formula for the determinant of a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is:\[\det(A) = ad - bc\]In our problem:- Matrix \(A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix}\)- Determinant \(\det(A) = 3 \times 9 - 4 \times 7 = 27 - 28 = -1\)A non-zero determinant, such as \(-1\) in this case, indicates that the matrix has an inverse, allowing us to proceed with solving the system using matrix inversion.
Verification of Solution
Once a solution is found by solving a matrix equation, it is essential to verify the correctness by substituting the solution back into the original equations.For the system:
- Solution obtained: \(x = -10\), \(y = 10\)
- First equation: \(3(-10) + 4(10) = -30 + 40 = 10\)
- Second equation: \(7(-10) + 9(10) = -70 + 90 = 20\)