Understanding Linear Algebra - Part 1

Linear algebra is the foundation of many fields, from computer graphics to machine learning. In this post, we will explore the row picture, column picture, and matrix form of a system of equations. We’ll also dive into the elimination method with detailed examples.

Row Picture, Column Picture, and Matrix Form

A system of linear equations can be visualized in three ways:

Row Picture: Each equation is represented as a line in a 2D plane.
Column Picture: The system is expressed as a combination of column vectors.
Matrix Form: The system is written compactly as a matrix equation.

Example System of Equations

Consider the system:

\[\begin{aligned} x + 2y &= 5 \\ 3x + 4y &= 6 \end{aligned}\]

Row Picture

In the row picture, each equation is a line in the 2D plane. The solution is the intersection of these lines.

{
  "type": "line",
  "data": {
    "labels": [-10, -5, 0, 5, 10],
    "datasets": [
      {
        "label": "x + 2y = 5",
        "data": [
          { "x": -10, "y": 7.5 },
          { "x": 10, "y": -2.5 }
        ],
        "borderColor": "rgba(75,192,192,1)",
        "fill": false
      },
      {
        "label": "3x + 4y = 6",
        "data": [
          { "x": -10, "y": 9 },
          { "x": 10, "y": -6 }
        ],
        "borderColor": "rgba(255,99,132,1)",
        "fill": false
      }
    ]
  },
  "options": {
    "scales": {
      "x": { "type": "linear", "position": "bottom" },
      "y": { "type": "linear" }
    }
  }
}

Column Picture

In the column picture, the system is written as:

\[x \begin{bmatrix} 1 \\ 3 \end{bmatrix} + y \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}\]

This means we are looking for a linear combination of the column vectors that equals the right-hand side vector.

Matrix Form

The system can also be written compactly as:

\[\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}\]

Elimination Method

The elimination method transforms the system into an equivalent one that is easier to solve. The goal is to eliminate one variable to solve for the other.

Example 1: Solving the System

Start with the system: \(\begin{aligned} x + 2y &= 5 \\ 3x + 4y &= 6 \end{aligned}\)
Multiply the first equation by 3: \(3x + 6y = 15\)
Subtract the second equation: \((3x + 6y) - (3x + 4y) = 15 - 6 \\ 2y = 9 \implies y = 4.5\)
Substitute ( y = 4.5 ) into the first equation: \(x + 2(4.5) = 5 \implies x = -4\)

The solution is ( x = -4, y = 4.5 ).

Example 2: Another System

Solve the system:

\[\begin{aligned} 2x + y &= 8 \\ x - y &= 2 \end{aligned}\]

Add the equations: \((2x + y) + (x - y) = 8 + 2 \\ 3x = 10 \implies x = \frac{10}{3}\)
Substitute ( x = \frac{10}{3} ) into the second equation: \(\frac{10}{3} - y = 2 \implies y = \frac{10}{3} - 6 \implies y = -\frac{8}{3}\)

The solution is ( x = \frac{10}{3}, y = -\frac{8}{3} ).

In the next part, we will explore determinants, inverse matrices, and their applications in solving systems of equations.