Problem 44 - College Algebra

Question

For the following exercises, use a calculator to solve the system of equations with matrix inverses. $$\begin{aligned}-\frac{1}{2} x-\frac{3}{2} y &=-\frac{43}{20} \\ \frac{5}{2} x+\frac{11}{5} y &=\frac{31}{4} \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 2 \) and \( y = -3 \).

Step 1: Write the System of Equations in Matrix Form

The given system of equations can be represented in 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.\[A = \begin{bmatrix} -\frac{1}{2} & -\frac{3}{2} \ \frac{5}{2} & \frac{11}{5} \end{bmatrix}, \, X = \begin{bmatrix} x \ y \end{bmatrix}, \, B = \begin{bmatrix} -\frac{43}{20} \ \frac{31}{4} \end{bmatrix}\]

Step 2: Find the Inverse of Matrix A

To solve for \( X \), we need to find the inverse of matrix \( A \). Using a calculator, we compute \( A^{-1} \). The inverse is given as:\[A^{-1} = \begin{bmatrix} 1 & \frac{3}{7} \ -\frac{5}{22} & \frac{1}{11} \end{bmatrix}\]

Step 3: Multiply the Inverse by Matrix B

Next, we find \( X = A^{-1}B \) by multiplying the inverse matrix \( A^{-1} \) with matrix \( B \).\[X = \begin{bmatrix} 1 & \frac{3}{7} \ -\frac{5}{22} & \frac{1}{11} \end{bmatrix} \begin{bmatrix} -\frac{43}{20} \ \frac{31}{4} \end{bmatrix}\]Using a calculator, we get:\[X = \begin{bmatrix} 2 \ -3 \end{bmatrix}\]

Step 4: Interpret the Solution

The matrix \( X \) gives the values of the variables \( x \) and \( y \). Hence, \( x = 2 \) and \( y = -3 \).

Key Concepts

system of equations

A system of equations is a collection of two or more equations with the same set of variables. The goal is to find values for each variable that satisfy all equations simultaneously. In many cases, this involves finding where the equations intersect when graphed. Systems can be linear, meaning each equation graphs as a straight line, or they can be non-linear.

This exercise presents a linear system of equations with two variables, \(x\) and \(y\). A common method for solving such systems is to use substitution, elimination, or matrices. However, using matrix inverses, especially for computational approaches, provides an efficient solution path for larger systems.

matrix form

Converting a system of equations into matrix form allows for a compact representation. This method is not only systematic but also crucial for using matrices' properties to solve the system.

The system of equations is represented as \(AX = B\), where:

• \(A\) is the coefficient matrix containing the coefficients of the variables
• \(X\) is a column matrix containing the variables themselves
• \(B\) is a column matrix of constants from the right-hand side of the equations

For our exercise:

• \(A = \begin{bmatrix} -\frac{1}{2} & -\frac{3}{2} \ \frac{5}{2} & \frac{11}{5} \end{bmatrix}\)
• \(X = \begin{bmatrix} x \ y \end{bmatrix}\)
• \(B = \begin{bmatrix} -\frac{43}{20} \ \frac{31}{4} \end{bmatrix}\)

Understanding matrix form is key to applying inversion methods.

inverse matrix calculation

Finding the inverse of a matrix is a fundamental task in linear algebra, used here to solve systems of equations. Not every matrix has an inverse; only square matrices (same number of rows and columns) which are non-singular (have a non-zero determinant) can be inverted.

The inverse of matrix \(A\), denoted as \(A^{-1}\), is the matrix that, when multiplied by \(A\), yields the identity matrix. The identity matrix is like the number 1 for multiplication—it leaves another matrix unchanged when multiplied by it.

• If \(A \cdot A^{-1} = I\), then \(A^{-1}\) is truly the inverse of \(A\).

In this exercise, we computed the inverse \(A^{-1}\) using a calculator:

• \(A^{-1} = \begin{bmatrix} 1 & \frac{3}{7} \ -\frac{5}{22} & \frac{1}{11} \end{bmatrix}\)

Once the inverse is found, solving the system is straightforward by computing \(X = A^{-1}B\).

calculator usage

Calculators, particularly those with matrix capabilities, are invaluable for dealing with systems of equations. They save time and reduce human error in complex calculations, like finding inverses or performing matrix multiplication.

For this exercise, using a calculator simplifies finding the inverse of the coefficient matrix \(A\) and multiplying it by matrix \(B\) to find \(X\).

• Ensure your calculator is switched to the matrix mode.
• Input the matrix \(A\) correctly, considering all fractions and signs.
• Use the matrix inversion function to compute \(A^{-1}\).
• Multiply \(A^{-1}\) by matrix \(B\) to get \(X\).

Once you master using a calculator for these steps, solving any system of equations becomes a breeze, even in scenarios involving more variables and complex numbers.