Problem 41 - Intermediate Algebra

Question

Use matrices to solve each system of equations. If the equations of a system are dependent or if a system is inconsistent, state this. $$ \left\\{\begin{array}{l} 3 x+4 y=-12 \\ 9 x-2 y=6 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = -1 \) and \( y = -3 \).

Step 1: Write the System in Matrix Form

The system of equations can be expressed in the matrix form as \( AX = B \), where \( A \) is the matrix of coefficients, \( X \) is the column matrix of variables, and \( B \) is the column matrix of constants. For the given system:\[A = \begin{bmatrix} 3 & 4 \ 9 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} -12 \ 6 \end{bmatrix}\]

Step 2: Check the Determinant

Calculate the determinant of matrix \( A \). If the determinant is zero, the system is either dependent or inconsistent.The determinant \( \text{det}(A) \) is calculated as:\[\text{det}(A) = (3)(-2) - (9)(4) = -6 - 36 = -42\]Since the determinant is non-zero (\(-42\)), the system has a unique solution.

Step 3: Find the Inverse of Matrix A

Since the determinant is non-zero, find the inverse of matrix \( A \). The inverse \( A^{-1} \) is given by:\[A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} -2 & -4 \ -9 & 3 \end{bmatrix}\]Plugging in the values, we get:\[A^{-1} = \frac{1}{-42} \begin{bmatrix} -2 & -4 \ -9 & 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{21} & \frac{2}{21} \ \frac{3}{14} & -\frac{1}{14} \end{bmatrix}\]

Step 4: Solve for X

Use the inverse to solve \( X = A^{-1}B \).\[X = \begin{bmatrix} \frac{1}{21} & \frac{2}{21} \ \frac{3}{14} & -\frac{1}{14} \end{bmatrix} \begin{bmatrix} -12 \ 6 \end{bmatrix}\]Calculate the matrix multiplication:\[X = \begin{bmatrix} \left(\frac{1}{21} \cdot -12 + \frac{2}{21} \cdot 6\right) \ \left(\frac{3}{14} \cdot -12 - \frac{1}{14} \cdot 6\right) \end{bmatrix} = \begin{bmatrix} -1 \ -3 \end{bmatrix}\]

Step 5: Interpret the Results

The solution \( X = \begin{bmatrix} -1 \ -3 \end{bmatrix} \) represents the values of \( x \) and \( y \) that satisfy both equations. Therefore, \( x = -1 \) and \( y = -3 \). Given the non-zero determinant, this unique solution is confirmed to be correct.

Key Concepts

Matrix Equations

Matrix equations are a way to express a system of linear equations using matrices. In our given problem, we have two equations:

• \( 3x + 4y = -12 \)
• \( 9x - 2y = 6 \)

To convert these into a matrix equation, we identify the coefficients of the variables \( x \) and \( y \). This allows us to represent the equations in the form \( AX = B \):

• \( A \) is the matrix of coefficients: \( \begin{bmatrix} 3 & 4 \ 9 & -2 \end{bmatrix} \)
• \( X \) is the column matrix of variables: \( \begin{bmatrix} x \ y \end{bmatrix} \)
• \( B \) is the column matrix of constants: \( \begin{bmatrix} -12 \ 6 \end{bmatrix} \)

A matrix equation is beneficial as it allows us to use matrix operations to find solutions efficiently by leveraging techniques such as finding the inverse of a matrix or using row-reduction methods.

Determinant Calculation

The determinant of a matrix is a special number that can tell us a lot about the matrix itself. Specifically, for a 2x2 matrix, it helps us understand if the system of equations has a unique solution, no solution, or infinitely many solutions. In our example:For matrix \( A \), the determinant \( \text{det}(A) \) is calculated as:\[\text{det}(A) = (3)(-2) - (9)(4) = -6 - 36 = -42\]Since \( \text{det}(A) \) is not zero, we can conclude that the matrix \( A \) is invertible, which means that the system of equations has a unique solution. If the determinant were zero, it would indicate that the system is either dependent or inconsistent, meaning no unique solution is possible.

Matrix Inverse

The inverse of a matrix provides a powerful way to solve systems of equations algebraically. Only square matrices (same number of rows and columns) that have a nonzero determinant possess an inverse. Given matrix \( A \) from our example, since the determinant is \(-42\) (which is non-zero), it is invertible.To find \( A^{-1} \), or the inverse of \( A \), use the following formula for a 2x2 matrix:\[A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Substitute the values from matrix \( A \):\[A^{-1} = \frac{1}{-42} \begin{bmatrix} -2 & -4 \ -9 & 3 \end{bmatrix} = \begin{bmatrix} \frac{1}{21} & \frac{2}{21} \ \frac{3}{14} & -\frac{1}{14} \end{bmatrix}\]This inverse can be used to find the values of \( x \) and \( y \) that solve the system, proving how such operations facilitate solving equations that arise in various real-world applications.

Unique Solution

In systems of linear equations, a unique solution means that there is exactly one set of values for variables that satisfy all equations simultaneously. Determining whether a unique solution exists involves checking the determinant of the coefficient matrix:- If \( \text{det}(A) eq 0 \), then the system typically has a unique solution.- If \( \text{det}(A) = 0 \), the system has either no solutions or infinitely many solutions.In our case, the determinant is \(-42\), a non-zero value, indicating that our given system of equations has a unique solution. Solving for \( X \) using matrix inverse, we multiply \( A^{-1} \) and \( B \):\[X = A^{-1}B = \begin{bmatrix} \frac{1}{21} & \frac{2}{21} \ \frac{3}{14} & -\frac{1}{14} \end{bmatrix} \begin{bmatrix} -12 \ 6 \end{bmatrix} = \begin{bmatrix} -1 \ -3 \end{bmatrix}\]Thus, the values \( x = -1 \) and \( y = -3 \) form the unique solution that satisfies both equations, as was determined through these calculations.