Problem 34 - College Algebra

Question

For the following exercises, solve the system using the inverse of a \(2 \times 2\) matrix. $$\begin{aligned} \text { 4. } \frac{1}{2} x+\frac{1}{5} y &=-\frac{1}{4} \\\ \frac{1}{2} x-\frac{3}{5} y &=-\frac{9}{4} \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{3}{2}\) and \(y = \frac{25}{8}\).

Step 1: Write the System of Equations in Matrix Form

First, we need to express the given system of equations using matrices. The system \( \frac{1}{2}x + \frac{1}{5}y = -\frac{1}{4} \) and \( \frac{1}{2}x - \frac{3}{5}y = -\frac{9}{4} \) can be written in matrix form as: \[\begin{bmatrix} \frac{1}{2} & \frac{1}{5} \ \frac{1}{2} & -\frac{3}{5} \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix}= \begin{bmatrix} -\frac{1}{4} \ -\frac{9}{4} \end{bmatrix}.\]

Step 2: Find the Inverse of the Coefficient Matrix

The coefficient matrix is \( A = \begin{bmatrix} \frac{1}{2} & \frac{1}{5} \ \frac{1}{2} & -\frac{3}{5} \end{bmatrix} \). To find the inverse, calculate the determinant of \( A \), which is \[ \text{det}(A) = \left( \frac{1}{2} \right) \left( -\frac{3}{5} \right) - \left( \frac{1}{5} \right) \left( \frac{1}{2} \right).\]This simplifies to \[ -\frac{3}{10} - \frac{1}{10} = -\frac{4}{10} = -\frac{2}{5}.\]The inverse \( A^{-1} \) is given by \[ A^{-1} = \frac{1}{\text{det}(A)} \times \begin{bmatrix} -\frac{3}{5} & -\frac{1}{5} \ \frac{1}{2} & \frac{1}{2} \end{bmatrix}.\]

Step 3: Simplify the Inverse Matrix

Substitute the determinant to find the inverse:\[ A^{-1} = \frac{1}{-\frac{2}{5}} \begin{bmatrix} -\frac{3}{5} & -\frac{1}{5} \ \frac{1}{2} & \frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & \frac{1}{2} \ -\frac{5}{4} & -\frac{5}{4} \end{bmatrix}.\]

Step 4: Multiply the Inverse with the Constant Matrix

Now we solve for \( \begin{bmatrix} x \ y \end{bmatrix} \) by multiplying \( A^{-1} \) with the constant matrix on the right:\[\begin{bmatrix} x \ y \end{bmatrix} = A^{-1} \begin{bmatrix} -\frac{1}{4} \ -\frac{9}{4} \end{bmatrix}.\]Hence,\[ \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & \frac{1}{2} \ -\frac{5}{4} & -\frac{5}{4} \end{bmatrix} \begin{bmatrix} -\frac{1}{4} \ -\frac{9}{4} \end{bmatrix}.\]

Step 5: Perform Matrix Multiplication

Calculate the matrix multiplication:\[ x = \left( \frac{3}{2} \right)\left( -\frac{1}{4} \right) + \left( \frac{1}{2} \right) \left( -\frac{9}{4} \right) = -\frac{3}{8} - \frac{9}{8} = -\frac{12}{8} = -\frac{3}{2}, \]\[ y = \left( -\frac{5}{4} \right)\left( -\frac{1}{4} \right) + \left( -\frac{5}{4} \right) \left( -\frac{9}{4} \right) = \frac{5}{16} + \frac{45}{16} = \frac{50}{16} = \frac{25}{8}. \]

Step 6: Final Solution: Solution of the System

Finally, the solution to the system of equations is \( x = -\frac{3}{2} \) and \( y = \frac{25}{8} \).

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same set of unknowns. Solving them involves finding the values of the unknowns that satisfy all the equations simultaneously. These systems can appear intimidating at first glance, but they are fundamental in mathematics because they help us understand multiple relationships at once.

For example, consider the system of equations provided:

• \( \frac{1}{2}x + \frac{1}{5}y = -\frac{1}{4} \)
• \( \frac{1}{2}x - \frac{3}{5}y = -\frac{9}{4} \)

By solving this system, we're essentially finding the intersection point of two lines on a graph, where both these linear equations are simultaneously satisfied. This intersection point gives us the values of \( x \) and \( y \).

Matrix Form

Matrices provide a compact and organized way to handle systems of equations. Placing the system in matrix form can simplify the process of finding solutions, especially for larger systems.

In matrix form, our system of equations is written as:\[\begin{bmatrix} \frac{1}{2} & \frac{1}{5} \\frac{1}{2} & -\frac{3}{5} \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -\frac{1}{4} \ -\frac{9}{4} \end{bmatrix}\]Here, the first matrix is the **coefficient matrix** comprising the coefficients of variables \( x \) and \( y \). The second matrix contains the variables we're solving for, and the third matrix contains the constants from the equations.

Expressing a system in this form makes it easier to apply matrix operations such as inversion and multiplication to solve the system.

Inverse of a Matrix

The inverse of a matrix is a crucial concept when solving systems of equations. It's similar to how division works with numbers. If you have a matrix \( A \), its inverse \( A^{-1} \) satisfies the equation \( A A^{-1} = I \), where \( I \) is the identity matrix. This identity matrix acts like the number 1 in regular arithmetic.

To find the inverse of the coefficient matrix\[A = \begin{bmatrix} \frac{1}{2} & \frac{1}{5} \\frac{1}{2} & -\frac{3}{5} \end{bmatrix},\]you need to calculate the determinant and then follow specific steps to determine \( A^{-1} \). In this example, after finding the determinant as \(-\frac{2}{5}\), the matrix inverse is found via specific computation, resulting in:\[A^{-1} = \begin{bmatrix} \frac{3}{2} & \frac{1}{2} \-\frac{5}{4} & -\frac{5}{4} \end{bmatrix}\].

Using the inverse matrix is an efficient method when dealing with larger systems or when computational tools are used for solving multiple matrices.

Matrix Multiplication

Matrix multiplication is a process applied to solve the matrix equations once we have the inverse matrix. In our situation, you multiply the inverse of the coefficient matrix (\( A^{-1} \)) by the constant matrix to find the values of \( x \) and \( y \).

The matrix expression is:\[\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} \frac{3}{2} & \frac{1}{2} \-\frac{5}{4} & -\frac{5}{4} \end{bmatrix} \begin{bmatrix} -\frac{1}{4} \-\frac{9}{4} \end{bmatrix}\]Performing the multiplication involves:

• Calculating each element in the resulting vector \( x \) and \( y \) by summing the products of corresponding entries.
• The calculations yield \( x = -\frac{3}{2} \) and \( y = \frac{25}{8} \), which are the solutions to the system.

Matrix multiplication is not commutative, which means that the order in which you multiply matters. Always maintain the order in operations especially when using matrices in solving systems of equations.