Problem 56 - Precalculus with Limits

Question

In Exercises 51-58, use an inverse matrix to solve (if possible) the system of linear equations. \(\begin{cases} \frac{5}{6}x - y = -20 \\ \frac{4}{3}x - \frac{7}{2}y = -51 \end{cases}\)

Step-by-Step Solution

Answer

The solution of the system of equations will be the calculated values of \(x\) and \(y\) in step 3. Verification of the result can be done by checking these values in the original equations given in the exercise, as shown in step 4.

Step 1: Write the system as a matrix equation

We can re-write the given system of linear equations:\[\begin{cases}\frac{5}{6}x - y = -20\\frac{4}{3}x - \frac{7}{2}y = -51\end{cases}\]as a matrix equation:\[\begin{bmatrix}\frac{5}{6} & -1\\frac{4}{3} & -\frac{7}{2}\end{bmatrix}\cdot\begin{bmatrix}x \y\end{bmatrix}=\begin{bmatrix}-20 \-51\end{bmatrix}\]

Step 2: Find the inverse of the coefficient matrix

Next, find the inverse of the coefficient matrix. The formula to find the inverse of a \(2 \times 2\) matrix \[A = \begin{bmatrix}a & b \c & d\end{bmatrix}\]is\[A^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b \-c & a\end{bmatrix}\].Using this formula, we get:\[A^{-1} = \frac{1}{(\frac{5}{6} \times -\frac{7}{2}) - (-1 \times \frac{4}{3})} \begin{bmatrix}-\frac{7}{2} & 1 \-\frac{4}{3} & \frac{5}{6}\end{bmatrix}\]

Step 3: Apply the inverse to solve for variables

Now, multiply the inverse of the coefficient matrix we got from the previous step with the matrix on the right side of the initial matrix equation. This gives the solution to the original system of linear equations:\[\begin{bmatrix}x \y\end{bmatrix}= \begin{bmatrix}-\frac{7}{2} & 1 \-\frac{4}{3} & \frac{5}{6}\end{bmatrix}\cdot\begin{bmatrix}-20 \-51\end{bmatrix}\]Solve this to finally get the values of \(x\) and \(y\).

Step 4: Check the solution

After finding the values of \(x\) and \(y\), substitute them back into the original system of linear equations to confirm that both equations hold true. If true, then the solution is correct.

Key Concepts

Matrix Equations

A matrix equation can be an efficient way to represent and solve a system of linear equations. In the context of the original exercise, the problem was expressed as two linear equations, which were then transformed into a matrix equation. This is a simple yet powerful tool.

Every matrix equation can be written in the form:

• Coefficient matrix: a matrix containing the coefficients of the variables.
• Variable matrix: a column matrix containing the variables, typically written as \((x, y)\).
• Constant matrix: a column matrix containing the constant terms from the equations.

The primary advantage of using matrix equations is the straightforwardness in their manipulation, especially when solving systems using inverse matrices.

Linear Equations

Linear equations are the foundation for matrix equations and systems of equations. These are equations that make a straight line when graphed in two dimensions. In our original exercise, we're given a system of linear equations which are first-degree equations in terms of both variables \(x\) and \(y\).

Characteristics of linear equations include:

• They form straight-line graphs.
• They have constant coefficients.
• No variables are raised to a power other than one.

Linear equations are crucial, as they create the framework for more complex equations, and often solve real-world problems involving rates, costs, and mixtures.

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. When these equations are linear, our task often involves finding the single set of variable values that satisfy all equations in the system simultaneously.

To solve a system of linear equations, we generally have a few methodological options:

• Substitution method: solving one equation for a variable and substituting that value into another equation.
• Elimination method: adding or subtracting equations to eliminate one variable, making it easier to solve for others.
• Matrix method: particularly using inverse matrices when dealing in high complexity scenarios or large systems.

Understanding and mastering systems of equations help in tackling many mathematical and real-world challenges, such as optimization problems in economics and complex engineering tasks.