2.2 Homogeneous and Particular

Consider the system of two equations $2 x - y = 4$ and $x + y - 6 z = - 4$ . Chapter 1 tells us what to expect: two independent conditions on three unknowns leave a line of solutions. This section is about the structure of that line — a structure shared by the solution set of every linear system, of any size.

First replace the right-hand sides by zeros: $2 x - y = 0$ and $x + y - 6 z = 0$ . This is the homogeneous version of the system. It is guaranteed to be consistent, since $open parenthesis 0, 0, 0 close parenthesis$ works, and a short elimination shows its solutions are all multiples of a single vector: $t open parenthesis 1, 2, 0.5 close parenthesis$ . In the language of the previous section, the homogeneous solution set is a span — here, a line through the origin.

Now return to the original right-hand sides. One solution can be found by trial: $𝐩 equals open parenthesis 2, 0, 1 close parenthesis$ satisfies both equations (check: $2 \cdot 2 - 0 = 4$ and $2 + 0 - 6 = - 4$ ). We call it a particular solution — nothing special about it, just some point that works. The figure below shows both solution sets at once.

Two parallel straight lines in 3D: a blue line through the origin holding the homogeneous solutions, and a red line through the marked point (2, 0, 1) holding the full solution set. A short dashed gray segment connects the origin to that point, showing the translation by one particular solution. There are no sliders. — The blue line through the origin holds the solutions of the homogeneous system; the red line holds the solutions of the original system. The dashed segment is the translation by the particular solution $𝐩 equals open parenthesis 2, 0, 1 close parenthesis$ .

Explore the figure

Orbit until both lines are clearly visible. Confirm they are parallel — same direction vector, different anchor points.
The blue line passes through the origin; the red line does not. Which of the two systems could you have guessed is the homogeneous one, from the picture alone?
Follow the dashed segment from the origin to the red point. Every point of the red line is reached the same way: start on the blue line, then add $𝐩$ .
Would a different particular solution — any other point of the red line — have served equally well as the anchor?

One solution plus all homogeneous solutions

Here is the reasoning, and it never uses the specific numbers. Suppose $𝐩$ solves the system $A 𝐱 = 𝐛$ and $𝐡$ solves the homogeneous system $A 𝐱 equals 0$ . Then $A (𝐩 + 𝐡) = A 𝐩 + A 𝐡 = 𝐛 + 0 = 𝐛$ , so $𝐩 + 𝐡$ is again a solution. Conversely, if $𝐪$ is any solution, then $A (𝐪 - 𝐩) = 𝐛 - 𝐛 = 0$ , so $𝐪$ differs from $𝐩$ by a homogeneous solution. Both directions together say:

$solution set of A 𝐱 = 𝐛 = 𝐩 + {solutions of A 𝐱 = 0}$

Geometrically: the solution set of a consistent system is a translated copy of the homogeneous solution set — the same span, picked up and carried to $𝐩$ , exactly as the dashed segment in the figure shows. The homogeneous system contributes the shape and direction; the particular solution contributes only the location.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.