Steps to solve 2x2 linear systems with inverse matrices.

📚 Introduction to Solving 2x2 Linear Systems with Inverse Matrices

A system of linear equations can be represented in matrix form, and solving it using the inverse of a matrix is a powerful technique. This method is particularly useful when dealing with multiple systems with the same coefficients. Let's explore how to solve a 2x2 system using this approach.

📜 Historical Context

The concept of using matrices to solve linear systems has its roots in the work of mathematicians like Arthur Cayley in the 19th century. Matrix algebra provided a concise and efficient way to represent and manipulate linear equations, paving the way for solving complex systems using inverse matrices.

🔑 Key Principles

• 🔢 Matrix Representation: Represent the system of equations in the form $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix.
• ➕ Finding the Inverse: Calculate the inverse of the coefficient matrix $A$, denoted as $A^{-1}$. For a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is given by $A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, provided that $ad - bc \neq 0$.
• ➗ Solving for Variables: Multiply both sides of the equation $AX = B$ by $A^{-1}$ to get $X = A^{-1}B$. This gives the solution for the variables.

📝 Step-by-Step Guide

1. Step 1: Write the System in Matrix Form

Given a system of equations:

$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$

Represent it as $AX = B$:

$\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$

2. Step 2: Find the Inverse of Matrix A

For $A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}$, the inverse $A^{-1}$ is:

$A^{-1} = \frac{1}{a_1b_2 - a_2b_1} \begin{bmatrix} b_2 & -b_1 \\ -a_2 & a_1 \end{bmatrix}$

Ensure that $a_1b_2 - a_2b_1 \neq 0$ (the determinant is not zero).

3. Step 3: Multiply $A^{-1}$ by B

Multiply the inverse matrix $A^{-1}$ by the constant matrix $B$ to find the solution matrix $X$:

$X = A^{-1}B$

$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{a_1b_2 - a_2b_1} \begin{bmatrix} b_2 & -b_1 \\ -a_2 & a_1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$

4. Step 4: Solve for x and y

Perform the matrix multiplication to find the values of $x$ and $y$.

$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{b_2c_1 - b_1c_2}{a_1b_2 - a_2b_1} \\ \frac{-a_2c_1 + a_1c_2}{a_1b_2 - a_2b_1} \end{bmatrix}$

Thus, $x = \frac{b_2c_1 - b_1c_2}{a_1b_2 - a_2b_1}$ and $y = \frac{-a_2c_1 + a_1c_2}{a_1b_2 - a_2b_1}$.

🌍 Real-world Examples

Consider the system:

$\begin{cases} 2x + 3y = 8 \\ x - y = -1 \end{cases}$

1. Matrix Form:

$\begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ -1 \end{bmatrix}$

2. Inverse of A:

$A^{-1} = \frac{1}{(2)(-1) - (3)(1)} \begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{-5} \begin{bmatrix} -1 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ \frac{1}{5} & -\frac{2}{5} \end{bmatrix}$

3. Multiply $A^{-1}$ by B:

$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & \frac{3}{5} \\ \frac{1}{5} & -\frac{2}{5} \end{bmatrix} \begin{bmatrix} 8 \\ -1 \end{bmatrix} = \begin{bmatrix} \frac{8}{5} - \frac{3}{5} \\ \frac{8}{5} + \frac{2}{5} \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

4. Solution:

$x = 1$ and $y = 2$

💡 Conclusion

Solving 2x2 linear systems using inverse matrices provides a structured and efficient method, especially when dealing with multiple systems sharing the same coefficient matrix. Understanding the underlying principles and practicing with examples can make this technique a valuable tool in your mathematical toolkit.