# How do I know that this system of equations has infinitely many solutions?

\begin{cases} 2x + 4y - 2z = 0\\ \\3x + 5y = 1 \end{cases}

My book is using this as an example of a system of equations that has infinitely many solutions, but I want to know how we can know that just from looking at the equations?

One can write your system $A x = b$ as augmented matrix and bring it into row echelon form $$\left[ \begin{array}{rrr|r} 2 & 4 & -2 & 0 \\ 3 & 5 & 0 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 3 & 5 & 0 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 0 & -1 & 3 & 1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 2 & -1 & 0 \\ 0 & 1 & -3 & -1 \end{array} \right] \to \left[ \begin{array}{rrr|r} 1 & 0 & 5 & 2 \\ 0 & 1 & -3 & -1 \end{array} \right]$$ This translates back into $$x + 5z = 2 \\ y - 3z = -1$$ or $x = (2-5z, -1+3z, z)^T$, where $z \in \mathbb{R}$. So there are infinite many solutions.

From a geometric point of view, each equation defines an affine plane in $\mathbb{R}^3$, which is a plane, not necessarily including the origin. $$(2, 4, -2) \cdot (x,y,z) = 0 \\ (3, 5, 0) \cdot (x, y, z) = 1$$ The first plane has a normal vector $(2,4,-2)^T$ and includes the origin, the second plane has normal vector $(3,5,0)$ and is $1/\sqrt{3^2 + 5^2}$ away from the origin.

The solution of the system is the intersection of those two planes. And only the cases empty intersection, intersection is a line or intersection is a plane can happen. Here the intersection is a line.

The image shows the two intersecting planes, the intersection line, and the point $P$ which corresponds to the solution with $z = 0$.

Because there are three variables and only two equations.

If you imagine $x$, $y$ and $z$ as the coordinates of a three-dimensional space, then any one equation involving one or more of $x$, $y$ and $z$ determines a plane in that space. For instance: $$0x+0y+z=2$$ determines a plane which is parallel to the $x,y$ plane but two units above it.

One equation defines a plane. Two equations define two planes: and in geometry, two planes intersect in a line. Thus these two equations define a line, and you know this just because there are two of them.

A third equation would define another plane, and the intersection of that plane and the line you have already got would give you a point. You know this just because there are three equations.

Of course there are exceptions, which you will learn - they arise because the two planes might be parallel (in which case they do not intersect) or they might be the same (in which case their intersection is a plane, not a line). These geometrical facts correspond to simple algebraic facts about the equations.

• Thanks, that makes sense. Thinking about it as two planes intersecting in a line really helped. – dagny Jul 12 '16 at 7:22

In general, you can exactly determine the value of (at most) one unknown per linear equation. You have two equations, but three unknowns ($x, y, z$), and so you can (assuming they do not contradict eachother) determine two of the values in terms of the third, but the third variable will not be restricted and can take on any value. This is what we mean by infinitely many solutions.

This is not the way to reason about the general problem, but it can give you an intuition about what's going on.

You can easily check that $(-3,2,1)$ is a solution of the system, so the set of solutions is nonempty. Now, suppose $(a, b, c)$ is a solution. Then $$f(a, b, c) = (\frac{a}{3}, b + \frac{2a} {3}, c-\frac{2a} {3})$$ is a solution too. Plus, this application is injective (you can check this by hand too, or put it into matrix form and check that its determinant is nonzero), so the set of solutions cannot be finite.

Each equation describe some plane in $\mathbb{R}^3$. These equations are not proportional so these planes have to intersect. Two planes intersect by line so you have infinitely many solutions.