Problem 43 - College Algebra

Question

For the following exercises, use a calculator to solve the system of equations with matrix inverses. $$\begin{aligned} 2 x-y &=-3 \\\\-x+2 y &=2.3 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = -1.2333\) and \(y = 0.5333\).

Step 1: Write the system in matrix form

To solve the system using matrix inverses, we first need to express the system in matrix form. We can represent the system as:\[ \begin{bmatrix} 2 & -1 \-1 & 2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -3 \ 2.3 \end{bmatrix} \]This matrix equation is in the form \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constants matrix.

Step 2: Find the inverse of the coefficient matrix

The next step is to find 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 our coefficient matrix \(A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix}\), the determinant \(ad-bc = 2(2) - (-1)(-1) = 4 - 1 = 3\). Thus, the inverse is:\[ A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \ \frac{1}{3} & \frac{2}{3} \end{bmatrix} \]

Step 3: Multiply the inverse of the coefficient matrix by the constants matrix

Now, we need to multiply \(A^{-1}\) by matrix \(B\) to find \(X\):\[ X = A^{-1}B = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \ \frac{1}{3} & \frac{2}{3} \end{bmatrix} \begin{bmatrix} -3 \ 2.3 \end{bmatrix} \]Perform the matrix multiplication:\[ \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \ \frac{1}{3} & \frac{2}{3} \end{bmatrix} \begin{bmatrix} -3 \ 2.3 \end{bmatrix} = \begin{bmatrix} \frac{2}{3}(-3) + \frac{1}{3}(2.3) \ \frac{1}{3}(-3) + \frac{2}{3}(2.3) \end{bmatrix} \]Calculating each term separately, we obtain:First row: \( \frac{2}{3}(-3) + \frac{1}{3}(2.3) = -2 + 0.7667 = -1.2333 \)Second row: \( \frac{1}{3}(-3) + \frac{2}{3}(2.3) = -1 + 1.5333 = 0.5333 \)

Step 4: Write the solution

From the matrix multiplication, we have found \( X = \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} -1.2333 \ 0.5333 \end{bmatrix} \). Thus, the solution to the system of equations is:\( x = -1.2333 \) and \( y = 0.5333 \).

Key Concepts

System of Equations

A system of equations refers to a set of two or more equations that share the same variables. In this exercise, we have two equations with two variables, \( x \) and \( y \). The goal is to find values for \( x \) and \( y \) that satisfy both equations simultaneously. The given system is:

• \( 2x - y = -3 \)
• \(-x + 2y = 2.3 \)

These kinds of systems can represent real-world problems, where each equation models a relationship between the variables. Solving such systems can lead to finding the intersection points of lines or other curves represented by the equations. By solving the system, we ensure that both equations hold true for the values of \( x \) and \( y \) found.
There are several methods to solve systems of equations, such as substitution, elimination, and using matrices. Using matrices, particularly the inverse of a matrix, can be a powerful method to handle complex systems efficiently.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, used to combine matrices in a precise way. When multiplying matrices, the number of columns in the first matrix must match the number of rows in the second matrix. In this solution, matrix multiplication is used to combine the inverse of the coefficient matrix with the constants matrix to find the variables matrix.

• Consider two matrices, \( A \) and \( B \): if \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) and \( B = \begin{bmatrix} e \ f \end{bmatrix} \), then their product \( AB \) is calculated as:

\[AB = \begin{bmatrix} ae + bf \ ce + df \end{bmatrix}\]
For our problem, after finding the inverse matrix \( A^{-1} \), we multiplied it by the constants matrix \( B \):

• \( A^{-1} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \ \frac{1}{3} & \frac{2}{3} \end{bmatrix} \)
• \( B = \begin{bmatrix} -3 \ 2.3 \end{bmatrix} \)

Then perform the multiplication to get \( X \), the solution for \( x \) and \( y \): \[\begin{bmatrix} \frac{2}{3}(-3) + \frac{1}{3}(2.3) \ \frac{1}{3}(-3) + \frac{2}{3}(2.3) \end{bmatrix}\]
Matrix multiplication allows us to streamline the solution process and manage calculations in a systematic way.

Determinant Calculation

The determinant is a special number that can be calculated from a square matrix, giving us critical insights into the matrix's properties. For a 2x2 matrix, the determinant is calculated as: \[det(A) = ad - bc\]Here \( A \) is the matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \).
The determinant offers important information, such as whether a matrix is invertible. A matrix with a determinant of zero does not have an inverse.
In this exercise, our coefficient matrix is \( A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \).
Here, the determinant would be calculated as:

• \( det(A) = 2 \cdot 2 - (-1) \cdot (-1) = 4 - 1 = 3 \)

Since the determinant is non-zero, the matrix is invertible, which means we can find a solution to the system using this method.

2x2 Matrix Inversion

Matrix inversion involves finding another matrix, which when multiplied with the original matrix, yields the identity matrix. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]In this exercise, the matrix to invert is \( A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \). Its determinant \( ad-bc \) was found to be 3. Hence, the inverse of the matrix is:\[A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \ \frac{1}{3} & \frac{2}{3} \end{bmatrix}\]
Finding the inverse is vital for solving systems of equations using matrices. By multiplying the inverse by the constants matrix, we can derive the solution matrix, which contains the values of the variables. Understanding how to find and apply inverses helps greatly in solving linear algebra problems efficiently.