2.2 Homogeneous and Particular

Consider the system of two equations $2x - y = 4$ and $x + y - 6z = -4$ . 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: $2x - y = 0$ and $x + y - 6z = 0$ . This is the homogeneous version of the system, in the sense of Section 1.4. It is guaranteed to be consistent, since $(0, 0, 0)$ works, and a short elimination shows its solutions are all multiples of a single vector: $t\,(1, 2, 0.5)$ . 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: $\mathbf{p} = (2, 0, 1)$ satisfies both equations (check: $2 \cdot 2 - 0 = 4$ and $2 + 0 - 6 = -4$ ). It is a particular solution in the sense of Section 1.4 — 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 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 gray segment is the translation by the particular solution $\mathbf{p} = (2, 0, 1)$ .

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 gray 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 $\mathbf{p}$ .
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. Write the system as $A\mathbf{x} = \mathbf{b}$ , shorthand that Section 2.4 makes official: $A$ is the matrix of coefficients, $A\mathbf{x}$ is the list of left-hand sides evaluated at the point $\mathbf{x}$ , and $\mathbf{b}$ is the list of right-hand sides. Left-hand sides of linear equations add and scale term by term, so $A(\mathbf{x} + \mathbf{y}) = A\mathbf{x} + A\mathbf{y}$ and $A(c\mathbf{x}) = cA\mathbf{x}$ ; that is all the argument uses. Suppose $\mathbf{p}$ solves the system $A\mathbf{x} = \mathbf{b}$ and $\mathbf{h}$ solves the homogeneous system $A\mathbf{x} = \mathbf{0}$ . Then $A(\mathbf{p} + \mathbf{h}) = A\mathbf{p} + A\mathbf{h} = \mathbf{b} + \mathbf{0} = \mathbf{b}$ , so $\mathbf{p} + \mathbf{h}$ is again a solution. Conversely, if $\mathbf{q}$ is any solution, then $A(\mathbf{q} - \mathbf{p}) = \mathbf{b} - \mathbf{b} = \mathbf{0}$ , so $\mathbf{q}$ differs from $\mathbf{p}$ by a homogeneous solution. Both directions together say:

$\text{solution set of } A\mathbf{x} = \mathbf{b} \; = \; \mathbf{p} + \{\text{solutions of } A\mathbf{x} = \mathbf{0}\}$

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 $\mathbf{p}$ , exactly as the gray segment in the figure shows. The homogeneous system contributes the shape and direction; the particular solution contributes only the location.

\begin{bmatrix}2 \\ 4 \\ 1\end{bmatrix}

The homogeneous solution set, as a null space ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

"A short elimination shows its solutions are all multiples of a single vector" is the one step above that the section asks you to take on faith. Here it is, done by machine: the null space of the coefficient matrix, which is precisely the solution set of $A\mathbf{x} = \mathbf{0}$ .

Worth noticing what the check had to tolerate. The engine returns the basis vector $(2, 4, 1)$ ; the text states $(1, 2, 0.5)$ . Those are different vectors spanning the same line. The null space itself is a fact — one definite subspace, here that line — but a basis for it is a choice, and the two choices differ. The comparison is therefore made by span rather than entry-by-entry, so the text keeps the scaling it chose for readability and still gets a certificate. An entrywise check would have called a correct book wrong.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Chapters 1–6 follow the topic sequence of Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL). Chapter 0's scope (its topics, their order, and the idea each figure shows) follows the Vectors chapter of Ximera's Linear Algebra: An Interactive Introduction. The prose, worked examples and figures were written for this book, which is not affiliated with or endorsed by the authors of either work. License: CC-BY-NC-SA-4.0.