A2.1-SystemsOfLE.rmd

## Systems of Linear Equations



Systems of linear equations are fundamental in linear algebra, as many problems can be formulated in this way. Linear algebra provides the tools to solve these systems efficiently.


<div class="example">
A company produces products \( N_1, \ldots, N_n \) using resources \( R_1, \ldots, R_m \).  Each product \( N_j \) requires \( a_{ij} \) units of resource \( R_i \).  If \( b_i \) units of each resource \( R_i \) are available, then the total resources used must satisfy

\[
a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n = b_i, \quad i = 1, \ldots, m
\]

or, in general matrix form,

\[
A\mathbf{x} = \mathbf{b},
\]

where \( A = [a_{ij}] \in \mathbb{R}^{m \times n} \), \( \mathbf{x} \in \mathbb{R}^n \), and \( \mathbf{b} \in \mathbb{R}^m \).
</div>


<div class="example">
Suppose a factory produces two products: **P1** and **P2**.  
Let:

- \(x_1\) = units of P1 produced  
- \(x_2\) = units of P2 produced  

**Constraints:**

1. Labor: Each unit of P1 requires 2 hours, P2 requires 3 hours, and total available labor is 120 hours:  
   \[
   2x_1 + 3x_2 \leq 120
   \]  
2. Material: Each unit of P1 uses 1 kg of material, P2 uses 2 kg, and total available material is 100 kg:  
   \[
   1x_1 + 2x_2 \leq 100
   \]  
3. Non-negativity:  
   \[
   x_1 \geq 0, \quad x_2 \geq 0
   \]

We can rewrite the inequalities as a matrix inequality:
\[
\underbrace{
\begin{pmatrix}
2 & 3 \\
1 & 2
\end{pmatrix}}_{A}
\begin{pmatrix}
x_1 \\ x_2
\end{pmatrix}
\leq
\underbrace{
\begin{pmatrix}
120 \\ 100
\end{pmatrix}}_{b}
\]

Here:

- \(A\) is the **constraint matrix**.  
- \(x = \begin{pmatrix}x_1 \\ x_2\end{pmatrix}\) is the **decision variable vector**.  
- \(b\) is the **resource vector**.  
</div>


---


###  Solutions to Systems of Linear Equations


In general, a system of linear equations can have:

- no solution,  
- exactly one solution, or  
- infinitely many solutions.  

<div class="example">
The system
\[
\begin{aligned}
x_1 + x_2 + x_3 &= 3 \\
x_1 - x_2 + 2x_3 &= 2 \\
2x_1 + 3x_3 &= 1
\end{aligned}
\]
has **no solution**, since combining the first two equations gives \( 2x_1 + 3x_3 = 5 \), which contradicts the third equation.  
</div>

Systems of equations with no solutions are called *inconsistent*.

<div class="definition">
A system of linear equations is said to be **inconsistent** if no set of values for the unknown variables satisfies all equations simultaneously.   
</div>

In other words, the equations contradict each other, and there is no common solution. 


<div class="example">
The system
\[
\begin{aligned}
x_1 + x_2 + x_3 &= 3 \\
x_1 - x_2 + 2x_3 &= 2 \\
x_2 + x_3 &= 2
\end{aligned}
\]
has **a unique solution** \( (x_1, x_2, x_3) = (1, 1, 1) \).
</div>


<div class="example">
The system
\[
\begin{aligned}
x_1 + x_2 + x_3 &= 3 \\
x_1 - x_2 + 2x_3 &= 2 \\
2x_1 + 3x_3 &= 5
\end{aligned}
\]
has **infinitely many solutions**, since the third equation is a linear combination of the first two.  
If we set \( x_3 = a \in \mathbb{R} \), then
\[
(x_1, x_2, x_3) =
\left( \tfrac{5}{2} - \tfrac{3}{2}a,\,
\tfrac{1}{2} + \tfrac{1}{2}a,\,
a \right),
\quad a \in \mathbb{R}.
\]
Hence, the solution set forms a line in \(\mathbb{R}^3\).
</div>




<div class="definition">
A system of linear equations is called **consistent** if there exists at least one set of values for the unknown variables that satisfies all equations simultaneously.  

- If there is exactly one solution, the system is **uniquely consistent**.  
- If there are infinitely many solutions, the system is **dependent** (still consistent).  
</div>


<div class="example">
The system of equations\[
\begin{cases}
x + y = 5 \\
2x - y = 1
\end{cases}
\]
is a consistent system with a unique solution: \(x = 2\), \(y = 3\).
</div>


<div class="example">
The system of equations 
\[
\begin{cases}
x + y = 4 \\
2x + 2y = 10
\end{cases}
\]
is inconsistent since the second equation is equivalent to \(x + y = 5\), which contradicts the first equation (\(x + y = 4\)).  Thus, the system has no solution.
</div>


---

### Geometric Interpretation

In two dimensions, each linear equation represents a **line** on the \( x_1x_2 \)-plane. The **solution set** is the intersection of these lines, which can be:

- a **point** (unique solution),  
- a **line** (infinitely many solutions), or  
- **empty** (no solution).  

For three variables, each equation defines a **plane** in \( \mathbb{R}^3 \).  Their intersection can be a plane, line, point, or empty set.


<div class="example">
\[
\begin{aligned}
4x_1 + 4x_2 &= 5 \\
2x_1 - 4x_2 &= 1
\end{aligned}
\]
has the unique solution \( (x_1, x_2) = (1, \tfrac{1}{4}) \).
</div>


<div class="example">
Consider the system:
\[
\begin{cases}
x + y + z = 3 \\
2x + 2y + 2z = 6 \\
x - y + z = 1
\end{cases}
\]
The second equation is just \(2 \times\) the first equation, so it doesn't add a new constraint.  The first and third equations define a plane intersection.  This leaves one free variable, so there are infinitely many solutions.


Let \(z = t\) (free parameter), then:
\[
\begin{aligned}
x + y + t &= 3 \implies x = 3 - y - t \\
x - y + t &= 1 \implies (3 - y - t) - y + t = 1 \implies 2y = 2 \implies y = 1 \\
x &= 3 - 1 - t = 2 - t
\end{aligned}
\]
Thus, the general solution is:
\[
(x, y, z) = (2 - t, 1, t), \quad t \in \mathbb{R}.
\]
</div>



---

### Matrix Formulation

A system of linear equations can be written compactly as:

\[
\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{bmatrix}
\begin{bmatrix}
x_1 \\ \vdots \\ x_n
\end{bmatrix}
=
\begin{bmatrix}
b_1 \\ \vdots \\ b_m
\end{bmatrix}
\]

This matrix representation \( A\mathbf{x} = \mathbf{b} \) provides a compact and powerful way to describe and solve systems of linear equations.



<div class="example">
Write the system of equations as a matrix: \[
\begin{cases}
x + y = 5 \\
2x - y = 1
\end{cases}
\]

The matrix version of this system is:
\[
\begin{bmatrix}
1 & 1 \\
2 & -1
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
\begin{bmatrix}
5 \\ 1
\end{bmatrix}
\]
</div>


<div class="example">
Write the system of equations as a matrix: \[\begin{cases}
x + y = 4 \\
2x + 2y = 10
\end{cases}
\]

The matrix version of this system is:
\[
\begin{bmatrix}
1 & 1 \\
2 & 2
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
\begin{bmatrix}
4 \\ 10
\end{bmatrix}
\]
</div>



---


### Exercises {.unnumbered .unlisted}



<div class="exercise">
Write a system of equations with a unique solution.

</div>


<div class="exercise">
Write a system of equations with infinitely many solutions. 
</div>


<div class="exercise">
Write a system of equations with no solutions. 
</div>



<div class="exercise">
Solve example 2.11

<div style="text-align: right;">
[Solution]( )
</div> 
</div>



<div class="exercise">
Write 2.15 and 2.16 as a matrix equations

<div style="text-align: right;">
[Solution]( )
</div> 
</div>




<div class="exercise">
Solve a system of equations

<div style="text-align: right;">
[Solution]( )
</div> 
</div>




<div class="exercise">
Solve a system of equations

<div style="text-align: right;">
[Solution]( )
</div> 
</div>


<div class="exercise">
Turn into a matrix form

<div style="text-align: right;">
[Solution]( )
</div> 
</div>




<div class="exercise">
Change from matrix to equations

<div style="text-align: right;">
[Solution]( )
</div> 
</div>

---