Problem 32 - College Algebra with Corequisite Support

Question

For the following exercises, solve the system using the inverse of a \(2 \times 2\) matrix. $$ \begin{array}{l} -2 x+3 y=\frac{3}{10} \\ -x+5 y=\frac{1}{2} \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = 0\) and \(y = \frac{1}{10}\).

Step 1: Identify Matrix Components

The first step is to identify the components needed to form the matrix equation. You have the system of equations:\(-2x + 3y = \frac{3}{10}\)\(-x + 5y = \frac{1}{2}\)We can represent this as a matrix equation \(AX = B\), where\[ A = \begin{pmatrix} -2 & 3 \ -1 & 5 \end{pmatrix}, \quad X = \begin{pmatrix} x \ y \end{pmatrix}, \quad B = \begin{pmatrix} \frac{3}{10} \ \frac{1}{2} \end{pmatrix} \]

Step 2: Find the Inverse of Matrix A

To solve for \(X\), we first need the inverse of \(A\) since \(X = A^{-1}B\). The inverse of a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by:\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]For our matrix \(A\), we have:\[ ad - bc = (-2)(5) - (3)(-1) = -10 + 3 = -7 \]Thus, the inverse of matrix \(A\) is:\[ A^{-1} = \frac{1}{-7} \begin{pmatrix} 5 & -3 \ 1 & -2 \end{pmatrix} = \begin{pmatrix} -\frac{5}{7} & \frac{3}{7} \ -\frac{1}{7} & \frac{2}{7} \end{pmatrix} \]

Step 3: Multiply Inverse with Matrix B

We now compute \(X = A^{-1}B\):\[ X = \begin{pmatrix} -\frac{5}{7} & \frac{3}{7} \ -\frac{1}{7} & \frac{2}{7} \end{pmatrix} \begin{pmatrix} \frac{3}{10} \ \frac{1}{2} \end{pmatrix} \]Calculate each component of \(X\):\[ x = -\frac{5}{7} \cdot \frac{3}{10} + \frac{3}{7} \cdot \frac{1}{2} = -\frac{15}{70} + \frac{3}{14} = -\frac{3}{14} + \frac{3}{14} = 0 \]\[ y = -\frac{1}{7} \cdot \frac{3}{10} + \frac{2}{7} \cdot \frac{1}{2} = -\frac{3}{70} + \frac{1}{7} = -\frac{3}{70} + \frac{10}{70} = \frac{7}{70} = \frac{1}{10} \]

Step 4: Solution Verification

Verify the solution \(x = 0\) and \(y = \frac{1}{10}\) by substituting back into the original equations:1. For \(-2x + 3y = \frac{3}{10}\): \(-2(0) + 3(\frac{1}{10}) = \frac{3}{10}\) This holds true.2. For \(-x + 5y = \frac{1}{2}\): \(-0 + 5(\frac{1}{10}) = \frac{5}{10} = \frac{1}{2}\) This also holds true, confirming our solution is correct.

Key Concepts

Understanding the 2x2 Matrix

A 2x2 matrix is a rectangular array of numbers with two rows and two columns. It is a simple yet powerful way to represent linear equations and transformations. For our specific problem, we have the matrix \( A = \begin{pmatrix} -2 & 3 \ -1 & 5 \end{pmatrix} \). This matrix contains coefficients from the system of equations that we are trying to solve.

Each row of the matrix corresponds to a different equation, and each column corresponds to a different variable. For example, the first row \( -2x + 3y = \text{something} \) corresponds to the first equation, while the second row \( -x + 5y = \text{something else} \) corresponds to the second equation. The 2x2 matrix is an efficient way to gather all important coefficients in one place, simplifying further computations.

Solving the System of Equations

A system of equations is a set of two or more equations that share variables. Our goal in solving a system is to find the specific values of these shared variables that satisfy all equations simultaneously. In this case, we need to find the values of \(x\) and \(y\) that make both equations true.

The given system is written as:

• \(-2x + 3y = \frac{3}{10}\)
• \(-x + 5y = \frac{1}{2}\)

By reformulating the problem into a matrix form, we can use matrix operations to efficiently find the solution. Methods like the Matrix Inverse Method offer a systematic route to tackle these simultaneous equations, particularly when directly manipulating individual equations becomes cumbersome.

The Importance of Solution Verification

Once we find a solution using the matrix method, it is essential to verify its correctness. Verification involves plugging the solution back into the original equations to confirm that it satisfies them. This step ensures that no mistakes were made during calculations.

Let's verify our solution:

• For the first equation \(-2x + 3y = \frac{3}{10}\), inputting our result \(x = 0\) and \(y = \frac{1}{10}\) gives \(-2(0) + 3\left(\frac{1}{10}\right) = \frac{3}{10}\). This is correct.
• For the second equation \(-x + 5y = \frac{1}{2}\), we substitute in to get \(-0 + 5\left(\frac{1}{10}\right) = \frac{1}{2}\). This too is correct.

Verification gives us confidence in the accuracy of our solution, demonstrating full consistency with the original problem.

Deciphering Matrix Multiplication

Matrix multiplication is a fundamental operation used in the Matrix Inverse Method to solve equations. It involves multiplying rows of one matrix with columns of another. In our exercise, this comes into play when calculating \(X = A^{-1}B\), where \(A^{-1}\) is the inverse matrix, and \(B\) is the constants' matrix.

For matrix elements:

• The element \(x\) is calculated as \(-\frac{5}{7} \times \frac{3}{10} + \frac{3}{7} \times \frac{1}{2}\), simplifying the process to eventually yield \(x = 0\).
• Similarly, \(y = -\frac{1}{7} \times \frac{3}{10} + \frac{2}{7} \times \frac{1}{2}\), resulting in \(y = \frac{1}{10}\).

Practicing matrix multiplication empowers students to solve complex problems efficiently and understand relationships between multiple variables within the matrix context.