On Systems & Row Reduction you learned to row reduce any system and read off its solutions. When there's exactly one solution, you're done: it's a list of numbers. But when there are infinitely many, you can't list them. You need a way to describe an infinite set, so that anyone can check whether a vector belongs to it, produce as many members as they like, and see its shape. This page is about that description, and about the surprisingly simple structure hiding underneath every solution set: one solution, plus every solution of a simpler system.
On Systems & Row Reduction you saw that three planes can meet in a point, a line, a plane, or nowhere. A point is easy to write down. But a line or a plane of solutions has infinitely many points, so you can't list them. You need a recipe for the whole set, one that lets anyone check whether a point is on it and walk to as many points as they like. This page is about that recipe, and about a picture hiding underneath every solution set: a line or plane through the origin, slid over to pass through one solution.
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away.
Here's the move, on a system that doesn't hand over its answer right away.
Find every solution of the system $$\begin{aligned} 2x + z &= 3 \\ x - y - z &= 1 \\ 3x - y &= 4. \end{aligned}$$
Put the equation with a plain $x$ on top ($R_1 \leftrightarrow R_2$), then clear $x$ below it ($R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$): $$\left(\begin{array}{rrr|r} 2 & 0 & 1 & 3 \\ 1 & -1 & -1 & 1 \\ 3 & -1 & 0 & 4 \end{array}\right) \to \left(\begin{array}{rrr|r} 1 & -1 & -1 & 1 \\ 0 & 2 & 3 & 1 \\ 0 & 2 & 3 & 1 \end{array}\right).$$ Rows 2 and 3 are now the same equation, so $R_3 \to R_3 - R_2$ turns row 3 into $0 = 0$. That leaves $2y + 3z = 1$ and $x - y - z = 1$, and column 3 has no pivot, so $z$ is free. Solving, $y = \frac{1}{2} - \frac{3}{2}z$ and $x = 1 + y + z = \frac{3}{2} - \frac{1}{2}z$: $$(x, y, z) = \left(\tfrac{3}{2}, \tfrac{1}{2}, 0\right) + z\left(-\tfrac{1}{2}, -\tfrac{3}{2}, 1\right).$$ The halves are honest but clumsy. Nothing says the free variable has to be the thing you vary: let $z = 1 + 2t$ instead, and every half cancels: $$(x, y, z) = (1, -1, 1) + t(-1, -3, 2).$$ Check $t = 0$: $(1, -1, 1)$ gives $2 + 1 = 3$, $1 + 1 - 1 = 1$, and $3 + 1 = 4$.
Both descriptions walk the same line, the one all three planes share. They just start in different places and walk at different speeds:
That $t$ deserves a name of its own, because it isn't quite the same thing as a free variable.
A parameter is a variable used to describe a family of solutions: as it runs through every real number, the description runs through every solution.
Free variables make natural parameters, and the reduced form hands them to you. But any description that lists the same set is just as correct.
Your general solution and the answer key's can look completely different and both be right. Different particular points and rescaled directions describe the same set. To compare two descriptions of a line, use the test from Lines & Planes: are the directions parallel, and is one starting point on the other line?
The free variable doesn't have to be the last one, either. Here it's in the middle of the list.
Find every solution of the system $$\begin{aligned} x + 2y &= 4 \\ y - z &= 0 \\ x + 2z &= 4. \end{aligned}$$
Row reduce with $R_3 \to R_3 - R_1$, then $R_3 \to R_3 + 2R_2$: $$\left(\begin{array}{rrr|r} 1 & 2 & 0 & 4 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 2 & 4 \end{array}\right) \to \left(\begin{array}{rrr|r} 1 & 2 & 0 & 4 \\ 0 & 1 & -1 & 0 \\ 0 & -2 & 2 & 0 \end{array}\right) \to \left(\begin{array}{rrr|r} 1 & 2 & 0 & 4 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right).$$ Row 2 says $y = z$ and row 1 says $x = 4 - 2y = 4 - 2z$, with $z$ free: $$(x, y, z) = (4 - 2z, \ z, \ z) = (4, 0, 0) + z(-2, 1, 1).$$ Check $z = 1$: $(2, 1, 1)$ gives $2 + 2 = 4$, $1 - 1 = 0$, and $2 + 2 = 4$.
In space, the third equation was never adding anything. Watch it swing until it says $0 = 0$ and drop out:
With two free variables, the description needs two parameters, one direction each.
Find every solution of the system $$\begin{aligned} x_1 + x_2 + x_3 - x_4 &= 1 \\ x_2 - x_3 + x_4 &= -1 \\ 3x_1 + 6x_3 - 6x_4 &= 6 \\ -x_2 + x_3 - x_4 &= 1. \end{aligned}$$
Clear below the first pivot with $R_3 \to R_3 - 3R_1$, which makes row 3 into $-3x_2 + 3x_3 - 3x_4 = 3$. Then $R_3 \to R_3 + 3R_2$ and $R_4 \to R_4 + R_2$ turn rows 3 and 4 into $0 = 0$, leaving $$\left(\begin{array}{rrrr|r} 1 & 1 & 1 & -1 & 1 \\ 0 & 1 & -1 & 1 & -1 \end{array}\right) \to \left(\begin{array}{rrrr|r} 1 & 0 & 2 & -2 & 2 \\ 0 & 1 & -1 & 1 & -1 \end{array}\right)$$ after $R_1 \to R_1 - R_2$. Columns 3 and 4 have no pivot, so $x_3 = s$ and $x_4 = t$ are free, $x_2 = -1 + s - t$, and $x_1 = 2 - 2s + 2t$: $$(x_1, x_2, x_3, x_4) = (2, -1, 0, 0) + s(-2, 1, 1, 0) + t(2, -1, 0, 1).$$
Find every solution of the system $$\begin{aligned} x - y + z &= 2 \\ 2x - 2y + 2z &= 4 \\ -x + y - z &= -2. \end{aligned}$$
$R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 + R_1$ turn rows 2 and 3 into $0 = 0$: all three equations were the same plane. That leaves $x - y + z = 2$, with $y = s$ and $z = t$ both free and $x = 2 + s - t$: $$(x, y, z) = (2, 0, 0) + s(1, 1, 0) + t(-1, 0, 1).$$ Check $s = 1$, $t = 1$: $(2, 1, 1)$ gives $2 - 1 + 1 = 2$.
Your turn: reduce, find the free variable, and pick the description that matches.
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A description is only useful if you can use it. The first thing to ask of one: is this particular vector in the set? You could plug the vector into the original equations. With the vector form you can also hunt for its parameter.
The first example's solutions are $(1, -1, 1) + t(-1, -3, 2)$. Is $(-1, -7, 5)$ one of them? Is $(3, 5, -2)$?
For $(-1, -7, 5)$, the first coordinate says $1 - t = -1$, so $t = 2$. Then the others have to agree: $-1 - 3(2) = -7$ and $1 + 2(2) = 5$. They do, so it's a solution.
For $(3, 5, -2)$, the first coordinate says $1 - t = 3$, so $t = -2$. The second agrees: $-1 - 3(-2) = 5$. But the third gives $1 + 2(-2) = -3$, not $-2$. One value of $t$ has to work in every coordinate at once, and none does: $(3, 5, -2)$ is not a solution.
That second vector is the trap: its first two coordinates agreed with $t = -2$, and only the last one gave it away.
Never stop after the first coordinate. One coordinate always gives you a value of $t$ (as long as the direction isn't $0$ there); the question is whether every other coordinate agrees with it.
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Every description so far had the same shape: one vector, plus free amounts of some directions. That's not luck. To see why, pair every system with a simpler one.
The associated homogeneous system of a system of linear equations is the system with the same coefficients and every right-hand side replaced by $0$.
A system and its homogeneous partner reduce the same way, move for move, because row operations only ever look at the coefficients to decide what to do. The right-hand sides just come along for the ride, and for the partner they ride along as zeros.
Solve $x + 2y = 5$, $3x - y = 1$ and its associated homogeneous system $x + 2y = 0$, $3x - y = 0$ side by side.
The same three moves work on both: $R_2 \to R_2 - 3R_1$, then $R_2 \to -\frac{1}{7}R_2$, then $R_1 \to R_1 - 2R_2$: $$\left(\begin{array}{rr|r} 1 & 2 & 5 \\ 3 & -1 & 1 \end{array}\right) \to \left(\begin{array}{rr|r} 1 & 2 & 5 \\ 0 & -7 & -14 \end{array}\right) \to \left(\begin{array}{rr|r} 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right),$$ $$\left(\begin{array}{rr|r} 1 & 2 & 0 \\ 3 & -1 & 0 \end{array}\right) \to \left(\begin{array}{rr|r} 1 & 2 & 0 \\ 0 & -7 & 0 \end{array}\right) \to \left(\begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right).$$ The system's solution is $(1, 2)$, and its partner's only solution is $(0, 0)$.
In pictures, each line has a partner through the origin, parallel to it, and every move swings the two alike:
Now name the one solution we'll build everything else from.
A particular solution of a system is any one solution of it.
"Any one" really means any. A system with infinitely many solutions has infinitely many particular solutions, and the theorem below works with whichever you pick. Here's the whole idea as a picture: the homogeneous solutions of $x - y + z = 0$ form a plane through the origin, and sliding that plane over by one solution $p$ of $x - y + z = 2$ gives every solution of $x - y + z = 2$.
Write equation $i$ of a system as $a_i \cdot x = d_i$, where $a_i$ is its row of coefficients and $x$ is the vector of unknowns: $2x - y + z$ is $(2, -1, 1) \cdot (x, y, z)$. The dot product splits over sums, $a \cdot (u + v) = a \cdot u + a \cdot v$, and that one rule does all the work below.
Let $p$ be a particular solution of a system of linear equations. Then the solutions of the system are exactly the vectors $p + h$, where $h$ is a solution of the associated homogeneous system.
Write equation $i$ as $a_i \cdot x = d_i$, so $a_i \cdot p = d_i$ for every $i$.
If $h$ solves the associated homogeneous system, then $a_i \cdot h = 0$ for every $i$, so $a_i \cdot (p + h) = a_i \cdot p + a_i \cdot h = d_i + 0 = d_i$. So $p + h$ solves the system.
Conversely, let $s$ be any solution of the system, and set $h = s - p$. Then $a_i \cdot h = a_i \cdot s - a_i \cdot p = d_i - d_i = 0$ for every $i$, so $h$ solves the associated homogeneous system, and $s = p + h$. $\blacksquare$
This is a visual argument, not a formal proof. Take a system's planes and its homogeneous partner's planes. Each pair has the same normal (the same coefficients), so each real plane is its partner slid off the origin, without turning. Here are $2x + z = 3$ and $x - y - z = 1$, starting from their partners through $0$:
The partners meet in a line through $0$; slid out, the real planes meet in a parallel line. Both directions of the theorem are one dot product each. A point $h$ on the partner line has $a \cdot h = 0$ for each normal $a$, so $a \cdot (p + h) = d + 0 = d$: $p + h$ is on every real plane. And if $s$ is on every real plane, then $a \cdot (s - p) = d - d = 0$, so $s - p$ is on the partner line, and $s = p + (s - p)$. The full proof is in the n version.
Remember it as the slogan: general = particular + homogeneous. The vector form you've been writing is exactly this, $p$ plus the homogeneous solutions $t(-1, -3, 2)$. It works for the other two kinds of solution set as well.
Solve $x + 2y - z = 1$, $2x + 4y = 2$, $y - 3z = 0$, and its associated homogeneous system.
$R_2 \to R_2 - 2R_1$ gives $2z = 0$, and swapping rows 2 and 3 leaves $$\left(\begin{array}{rrr|r} 1 & 2 & -1 & 1 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 2 & 0 \end{array}\right),$$ a pivot in every column. Back-substituting, $z = 0$, $y = 0$, $x = 1$: the one solution is $(1, 0, 0)$. The homogeneous partner reduces with the same moves to the same left side with zeros after the bar, so its only solution is $0$. General = particular + homogeneous = $(1, 0, 0) + 0$.
The empty kind fits too, in its own way: the homogeneous part is always there, but the particular solution might not be.
Solve $x + z = 1$, $y + z = 1$, $x + y + 2z = 3$, and its associated homogeneous system.
$R_3 \to R_3 - R_1 - R_2$ (two moves) turns row 3 into $0 = 1$: no solution at all. The homogeneous partner's row 3 becomes $0 = 0$ instead, leaving $x + z = 0$ and $y + z = 0$ with $z = t$ free: its solutions are $t(-1, -1, 1)$, a whole line. So the homogeneous part is infinite, but there is no particular solution to add it to, and the general solution is empty.
So the homogeneous system is the system's personality, and the right-hand side is just where it decided to stand today. Deep. Also slightly insulting to the right-hand side.
Practice the bookkeeping: build a general solution from its two parts, strip it back down, or move the right-hand sides.
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The same dot product trick answers a natural question: if you already know some solutions, which combinations of them are solutions too? Plugging $c_1s_1 + c_2s_2$ into one equation $a \cdot x = d$ gives $(c_1 + c_2)d$. So everything depends on what the weights add up to.
Let $s_1, \ldots, s_k$ be solutions of a system of linear equations, let $h_1, \ldots, h_\ell$ be solutions of its associated homogeneous system, and let $$x = c_1s_1 + \cdots + c_ks_k + e_1h_1 + \cdots + e_\ell h_\ell.$$ If $c_1 + \cdots + c_k = 1$, then $x$ solves the system. If $c_1 + \cdots + c_k = 0$, then $x$ solves the associated homogeneous system.
Take any equation $a \cdot x = d$ of the system. Each $s_j$ satisfies $a \cdot s_j = d$ and each $h_j$ satisfies $a \cdot h_j = 0$, so, because the dot product splits over sums and scalar multiples, $$a \cdot x = c_1(a \cdot s_1) + \cdots + c_k(a \cdot s_k) + e_1(a \cdot h_1) + \cdots + e_\ell(a \cdot h_\ell) = (c_1 + \cdots + c_k)d.$$ If the $c$'s add to $1$ this is $d$, for every equation, so $x$ solves the system. If they add to $0$ it is $0$, so $x$ solves the associated homogeneous system. $\blacksquare$
Let $s_1$ and $s_2$ be solutions of a system of linear equations, and let $h$ be a solution of its associated homogeneous system. Then $(1 - t)s_1 + ts_2 + h$ solves the system for every number $t$, and $s_1 - s_2$ solves the associated homogeneous system.
This is a visual argument, not a formal proof. The points $(1 - t)s_1 + ts_2$ are the straight line through $s_1$ and $s_2$, and on Systems & Row Reduction you saw that line stay inside every plane of the system. Adding $h$ is harmless: $a \cdot h = 0$ says $h$ is perpendicular to each normal, so it runs along every plane, and stepping by $h$ keeps a point on each plane. And $s_1 - s_2$ gives $d - d = 0$ in each equation. Here's the first example, with three combinations marked:
The weights on $s_1$ and $s_2$ decide where a combination lands: weights adding to $1$ stay on the system's line, weights adding to $0$ land on the homogeneous line through $0$, and anything else lands on neither. The full proof is in the n version.
Here it is with two solutions you've already met: $(1, -1, 1)$ from the first example, and $(0, -4, 3)$, which is its $t = 1$.
Both $s_1 = (1, -1, 1)$ and $s_2 = (0, -4, 3)$ solve the first example's system $2x + z = 3$, $x - y - z = 1$, $3x - y = 4$. Which of $2s_1 - s_2$, $s_1 + s_2$, and $s_1 - s_2$ solve it?
$2s_1 - s_2 = (2, 2, -1)$: its weights $2$ and $-1$ add to $1$, so it's a solution. Check: $4 - 1 = 3$, $2 - 2 + 1 = 1$, $6 - 2 = 4$.
$s_1 + s_2 = (1, -5, 4)$: weights add to $2$, and it gives $2 + 4 = 6$, $1 + 5 - 4 = 2$, $3 + 5 = 8$, every right-hand side doubled. Not a solution.
$s_1 - s_2 = (1, 3, -2)$: weights add to $0$, so it solves the homogeneous partner instead: $2 - 2 = 0$, $1 - 3 + 2 = 0$, $3 - 3 = 0$. It's $-1$ times the direction $(-1, -3, 2)$.
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General = particular + homogeneous splits the question "how many solutions?" into two smaller ones. Is there any solution at all? And if so, how many solutions does the homogeneous partner have? The second question only looks at the coefficients.
Suppose a system of linear equations has at least one solution. Then it has exactly one solution if its associated homogeneous system has only the trivial solution, and infinitely many solutions otherwise.
Let $p$ be a solution. By the theorem on general = particular + homogeneous, the solutions are exactly the vectors $p + h$ with $h$ a homogeneous solution. If the only homogeneous solution is $0$, the only solution is $p$. Otherwise there is a homogeneous solution $h \neq 0$, and every multiple $th$ is one too, since $a \cdot (th) = t(a \cdot h) = 0$ in each equation. Different values of $t$ give different vectors $p + th$ (subtract two of them: $(t - t')h \neq 0$ when $t \neq t'$), so there are infinitely many solutions. $\blacksquare$
This is a visual argument, not a formal proof. Changing the right-hand sides slides each plane without turning it, and the homogeneous partner is the version where every plane passes through $0$. Once there's one solution $p$, the solutions are $p$ plus the partner's solutions: a single point if the partner meets only at $0$, a copy of its line or plane otherwise. Here are two sets of left sides with their right-hand sides sliding the whole time:
The first set's partner meets only at $0$, and the slid planes always keep exactly one corner. The second set's partner shares a line, and the slid planes only ever make a prism (no solution) or share a whole line (infinitely many), never one point. The full proof is in the n version.
The "at least one solution" isn't fine print. Here's the theorem at work on two pairs of left sides.
For which right-hand sides does $x + 2y = a$, $3x + 4y = b$ have exactly one solution? And $x + 2y = a$, $3x + 6y = b$?
The first pair's homogeneous partner $x + 2y = 0$, $3x + 4y = 0$ reduces ($R_2 \to R_2 - 3R_1$ gives $-2y = 0$) to only $(0, 0)$. Reducing the real system the same way always gives a solution, $y = \frac{3a - b}{2}$ and then $x = b - 2a$, so there is exactly one solution for every $a$ and $b$.
The second pair's partner reduces to $x + 2y = 0$, $0 = 0$: a whole line of homogeneous solutions. So the real system never has exactly one solution. $R_2 \to R_2 - 3R_1$ turns its second row into $0 = b - 3a$: with $a = 1$, $b = 2$ that's $0 = -1$ and there's no solution; with $a = 1$, $b = 3$ it's $0 = 0$ and there are infinitely many.
Slide the right-hand sides yourself and watch which counts are even possible:
This also gives a second reason a system can't have exactly two solutions. A homogeneous system with one nonzero solution $h$ has all its multiples $th$, so its count is $1$ or infinity, and adding a particular solution keeps that count. Systems & Row Reduction got there by walking the line between two solutions; this way gets there by splitting the question in two.
"Infinitely many" is a fact about $\R$. Over the two-element field $\mathbb{F}_2 = \{0, 1\}$, a homogeneous solution $h \neq 0$ has only two multiples, $0$ and $h$, so a line of solutions has exactly $2$ points. There, a system's solution count is $0$ or a power of $2$: with $k$ free variables, exactly $2^k$.
Infinitely many solutions isn't a failure. In the real world it usually means you have choices, and the general solution is the menu.
A farmer has $1200$ acres to plant with corn, soybeans, and oats, and wants to use every acre. Seed costs $\$20$ an acre for corn, $\$50$ for soybeans, and $\$10$ for oats, and the farmer will spend exactly $\$30{,}000$ on seed. What are the possible plans?
Call the acres $c$, $s$, and $o$. Then $c + s + o = 1200$ and $20c + 50s + 10o = 30000$. Row reduce with $R_2 \to R_2 - 20R_1$, which gives $30s - 10o = 6000$, then $R_2 \to \frac{1}{30}R_2$ and $R_1 \to R_1 - R_2$: $$\left(\begin{array}{rrr|r} 1 & 0 & \frac{4}{3} & 1000 \\ 0 & 1 & -\frac{1}{3} & 200 \end{array}\right).$$ The oats $o$ are free, so $s = 200 + \frac{1}{3}o$ and $c = 1000 - \frac{4}{3}o$. Thirds are clumsy, so let $o = 300 + 3t$, which puts $t = 0$ at the even split: then $s = 300 + t$ and $c = 600 - 4t$, and the plans are $$(c, s, o) = (600, 300, 300) + t(-4, 1, 3).$$ Every value of $t$ solves the equations, but a field can't have negative acres. Corn stays at least $0$ while $t \le 150$, and oats while $t \ge -100$, so the real plans are $-100 \le t \le 150$: from $(1000, 200, 0)$, all corn and soy, to $(0, 450, 750)$, no corn at all.
If corn sells for $\$100$ an acre, soybeans for $\$300$, and oats for $\$80$, which plan is best? Each step of $t$ trades $4$ acres of corn for $1$ of soybeans and $3$ of oats, which changes the revenue by $-400 + 300 + 240 = 140$ dollars. So more $t$ is always better, and the best plan sits at the far end. Find it:
Writing out every equation is getting long. On the next page, Matrices, a whole system becomes one equation $Ax = b$, and its associated homogeneous system is $Ax = 0$. General = particular + homogeneous then reads: every solution of $Ax = b$ is $p + h$ with $Ah = 0$.
The homogeneous solutions $s(1, 1, 0) + t(-1, 0, 1)$ are every combination of two vectors. On Linear Combinations & Span that set gets a name, the span, and the homogeneous solutions of any system always form one.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Figuring out which idea a question needs is half the work.
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Made it this far? Time for something with a little heat. A parabola $y = ax^2 + bx + c$ through a point is one linear equation in $a$, $b$, and $c$. Two points give two equations in three unknowns, so there's a whole line of parabolas through them:
Write that family as particular + homogeneous, then pick out the one parabola that meets one more condition.
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