Problem 40 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System as a Matrix Equation Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix, as in Example \(6 .\) $$\left\\{\begin{array}{l}3 x+4 y=10 \\\7 x+9 y=20\end{array}\right.$$

Step-by-Step Solution

Answer

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

Step 1: Write the System as a Matrix Equation

To write the system of equations \(3x + 4y = 10\) and \(7x + 9y = 20\) as a matrix equation, identify the coefficient matrix \(A\), the variable matrix \(X\), and the constant matrix \(B\). The coefficient matrix \(A\) is \(\begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix}\), the variable matrix \(X\) is \(\begin{bmatrix} x \ y \end{bmatrix}\), and the constant matrix \(B\) is \(\begin{bmatrix} 10 \ 20 \end{bmatrix}\). Thus, the matrix equation is expressed as:\[ A \cdot X = B \] where \[ \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 10 \ 20 \end{bmatrix} \]

Step 2: Compute the Inverse of the Coefficient Matrix

To solve for \(X\), we first need the inverse of the coefficient matrix \(A\). The formula for the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is:\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]For \(A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix}\), calculate the determinant \(ad-bc\):\[ (3)(9) - (4)(7) = 27 - 28 = -1 \]Therefore, the inverse \(A^{-1}\) is:\[ A^{-1} = \frac{1}{-1} \begin{bmatrix} 9 & -4 \ -7 & 3 \end{bmatrix} = \begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix} \]

Step 3: Use the Inverse to Solve for the Variable Matrix

Using the inverse matrix \(A^{-1}\), solve for \(X\) by multiplying both sides of the equation \(A \cdot X = B\) by \(A^{-1}\):\[ A^{-1} \cdot A \cdot X = A^{-1} \cdot B \]Since \(A^{-1} \cdot A\) is the identity matrix \(I\), the equation simplifies to:\[ X = A^{-1} \cdot B \]Substitute \(A^{-1}\) and \(B\):\[ X = \begin{bmatrix} -9 & 4 \ 7 & -3 \end{bmatrix} \begin{bmatrix} 10 \ 20 \end{bmatrix} = \begin{bmatrix} (-9)(10) + (4)(20) \ (7)(10) + (-3)(20) \end{bmatrix} \]Simplify the multiplication:\[ X = \begin{bmatrix} -90 + 80 \ 70 - 60 \end{bmatrix} = \begin{bmatrix} -10 \ 10 \end{bmatrix} \]Thus, \(x = -10\) and \(y = 10\).

Step 4: Verify the Solution

To ensure our solution is correct, substitute \(x = -10\) and \(y = 10\) back into the original equations.For the first equation \(3x + 4y = 10\):\[ 3(-10) + 4(10) = -30 + 40 = 10 \text{ (Correct)} \]For the second equation \(7x + 9y = 20\):\[ 7(-10) + 9(10) = -70 + 90 = 20 \text{ (Correct)} \]Both equations hold true, confirming that \(x = -10\) and \(y = 10\) is the correct solution.

Key Concepts

Understanding Linear Systems

Linear systems are a collection of equations where each term is either a constant or the product of a constant and a single variable. These systems can consist of two or more linear equations that have the same set of variables. In the context of matrices, these systems permit us to employ matrix operations to find solutions efficiently, especially as the number of equations grows.

To approach solving a linear system using matrices, you first need three essential components: the coefficient matrix, the variable matrix, and the constant matrix.

• The coefficient matrix represents the coefficients of the variables in the system.
• The variable matrix contains the variables of the system.
• The constant matrix has the constant terms from each equation.

Together, these can be expressed as a matrix equation of the form: \( A \cdot X = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constant matrix. Solving the linear system, therefore, involves manipulating these matrices to find the values of the variables.

The Inverse of a Matrix

The inverse of a matrix plays a critical role in solving matrix equations, particularly when tackling linear systems. A matrix’s inverse, denoted \( A^{-1} \), is essentially its "flip side" such that when it's multiplied by the original matrix \( A \), it yields the identity matrix \( I \) (much as multiplying a number by its reciprocal yields 1).

Not all matrices have inverses, and the matrix must be square (the same number of rows and columns) and have a non-zero determinant to have an inverse. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse is calculated using the formula: \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \] Where \( ad-bc \) is the determinant of the matrix. This determinant must not be zero—if it is, the matrix doesn't have an inverse.

Finding the inverse is crucial for solving the equation \( A \cdot X = B \) because it allows conversion to \( X = A^{-1} \cdot B \), thereby giving us our desired variable matrix \( X \).

Exploring Determinants

Determinants are scalar values computed from a square matrix that provide essential information about the matrix, including whether it has an inverse. The determinant can effectively indicate the potential to solve a system of linear equations using matrices.

For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( ad-bc \). This value serves several purposes:

• If the determinant is zero, the matrix is said to be singular, meaning it does not have an inverse, and the system cannot be solved by that matrix method.
• A non-zero determinant signifies that the matrix is invertible, allowing us to find a unique solution to the matrix equation.

In our example exercise, the determinant was \( -1 \), confirming that the matrix is invertible. Thus, the matrix method can be confidently used to find the variables \( x \) and \( y \) in the linear system.