Systems of Linear Equations

A system of linear equations is a set of two or more equations involving the same variables that must be solved simultaneously.

\[
\begin{cases}
a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1 \\
a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2 \\
\vdots \\
a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = b_m
\end{cases}
\]

Here, \( a_1, a_2, \ldots, a_n \) are the coefficients of the variables \( x_1, x_2, \ldots, x_n \), and \( b_1, b_2, \ldots, b_n \) are constant terms.

Each equation in the system is linear, meaning it can be represented in the form:

The primary objective in a system of linear equations is to find the values of the variables that satisfy all the equations simultaneously.

There are several methods to solve these systems, including:

1. Substitution Method
   Solve one of the equations for one variable in terms of the others. Then substitute this expression into the remaining equations, reducing the number of variables. Repeat the process until all variables are solved.
2. Elimination Method (or Gaussian Elimination)
   Add or subtract equations to eliminate one of the variables. Repeat the process until a triangular (or echelon) form is achieved. Solve the resulting system by back-substitution.
3. Matrix Inversion Method
   Represent the system as a matrix \( \mathbf{A} \) of coefficients, a vector of variables \( \mathbf{x} \), and a vector of constants \( \mathbf{b} \). If \( \mathbf{A} \) is invertible, the solution can be found using \( \mathbf{x} = \mathbf{A}^{-1} \mathbf{b} \).

Practical Example

Let's consider a simple system of two linear equations with two variables:

\[
\begin{cases}
2x + 3y = 5 \\
4x - y = 3
\end{cases}
\]

We use the substitution method to solve the first equation for \( y \):

\[ y = \frac{5 - 2x}{3} \]

Substitute \( y \) into the second equation:

\[ 4x - \frac{5 - 2x}{3} = 3 \]

Solve for \( x \):

\[ 12x - 5 + 2x = 9 \]

\[ 14x = 14 \]

\[ x = 1 \]

Using the value of \( x \), solve for \( y \):

\[ y = \frac{5 - 2(1)}{3} = 1 \]

Thus, the solution to the system is \( x = 1 \) and \( y = 1 \).

In systems of equations with two or three variables, the symbols x, y, and z are typically used to denote the unknowns.

In systems with more variables, it's common to use the symbol x with indices (x₁, x₂, x₃, x₄...) to indicate the variables.

\[
\begin{cases}
a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1 \\
a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2 \\
\vdots \\
a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = b_m
\end{cases}
\]

A solution to the system \( S \) is a \( q \)-tuple of numbers \( (x_1, x_2, \ldots, x_q) \) that, when substituted into the system's equations for the unknowns \( (x_1, x_2, \ldots, x_q) \), satisfy the identities.

The coefficients (a₁, a₂, a₃, a₄...) of the variables form a matrix called the coefficient matrix.

\[
A = \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1q} \\
a_{21} & a_{22} & \cdots & a_{2q} \\
\vdots & \vdots & \ddots & \vdots \\
a_{p1} & a_{p2} & \cdots & a_{pq}
\end{pmatrix}
\]

This matrix is very useful for analyzing the solutions of the system of equations.

For example, let's take the previous example's system of equations and denote the unknowns as x₁ and x₂ instead of x and y.

\[
\begin{cases}
2x_1 + 3x_2 = 5 \\
4x_1 - x_2 = 3
\end{cases}
\]

In this case, the coefficient matrix \( A \) is a 2x2 square matrix.

\[  A = \begin{pmatrix}
2 & 3 \\
4 & -1
\end{pmatrix}
\]

The vector of unknowns consists of two elements x₁ and x₂.

$$ \vec{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} $$

The vector of constants consists of two numbers, which are the constant terms of the equations.

$$ \vec{b} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} $$

This allows the system of equations to be written in a different form, known as the matrix form.

$$ A \cdot \vec{x} = \vec{b} $$

$$ \begin{pmatrix}
2 & 3 \\
4 & -1
\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} $$

By performing matrix calculations, we obtain the system of equations again.

For example, let's multiply matrix \( A \) by vector \( \vec{x} \).

$$ \begin{pmatrix}
2x_1 & 3x_2 \\
4x_1 & -1x_2
\end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} $$

The final result is the original system of equations.

\[
\begin{cases}
2 x_1 + 3x_2 = 5 \\
4x_1 - x_2 = 3
\end{cases}
\]

In matrix form, this system can be efficiently solved using Cramer's rule.