Problem 24 - Precalculus: Mathematical for Calculus

Question

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. $$\left\\{\begin{array}{l}3 x+4 y=10 \\ 7 x+9 y=20\end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \( x = -10 \), \( y = 10 \).

Step 1: Write the System of Equations

The given system of equations is:\[\begin{align*}3x + 4y &= 10 \7x + 9y &= 20\end{align*}\]

Step 2: 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:\[ \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 10 \ 20 \end{bmatrix} \]

Step 3: 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:\[A^{-1} = \frac{1}{-1} \begin{bmatrix} 9 & -4 \ -7 & 3 \end{bmatrix} = \begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix}\]

Step 4: 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 \( y = 10 \).

Step 5: 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.

Key Concepts

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:

• \(3x + 4y = 10\)
• \(7x + 9y = 20\)

The idea is to express this system as a single matrix equation of the form \(A\mathbf{x} = \mathbf{b}\). Here, \(A\) is the matrix that contains the coefficients of the variables (\(x\) and \(y\)), \(\mathbf{x}\) is the matrix of variables, and \(\mathbf{b}\) includes the constants on the other side of the equation. In matrix terms, it looks like this:\[A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 10 \ 20 \end{bmatrix}\]Therefore, the system of equations is transformed into:\[\begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 10 \ 20 \end{bmatrix}\]

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: \(7(-10) + 9(10) = -70 + 90 = 20\)

Substituting these values confirms that both equations are satisfied, thus verifying our solution.Verification ensures that no mistakes were made during the calculation process and that the obtained result accurately solves the initial system of equations. This step is crucial for ensuring the reliability of your mathematical work.