Systems of Equations
The Geometry of three planes in space
&
the Reduced Row-echelon Form
Pat Ballew
Lakenheath, UK
Some background
Early in Geometry students learn that when two planes intersect, they intersect in a
single straight line. Few are taught the tools to determine the line of intersection, and
express it with an equation. This paper attempts to provide an approach to developing
the tools to accomplish that task and expand their understanding of the information
available in the use of the reduced row-echelon form of the augmented matrix of a system
of three equations in three unknowns.
We begin by recalling that an equation in three variables, such as 2x+3y+z=6 can
represent a plane in space. When students had three such equations that intersected in a
unique point, they found the solution by one of several methods. Most students learn to
solve such equations by the methods called elimination and substitution at the very least.
Others may have also been introduced to Cramer’s rule for solving systems with
determinants and perhaps two methods using matrices.
The most commonly taught matrix method is to write a matrix equation and then
solve it using the inverse matrix method. A second, and as we will point out, more
efficient and general method is the Gauss-Jordan reduced row-echelon form (RREF) of an
augmented matrix. We give an example of both below to clarify the terminoligy.
We begin with three planes determined by the equations {x + y – 2z = 9; 2x – 3y
+ z = -2; and x + 3y + z = 2} This same system of equations can be expressed as the
matrix equation
1 1
2
x

3 1
y
1 3
2
1
9
2
z
2
.
Notice that the left matrix is made up of the coefficients of the three variable terms
in each equation, and the right matrix contains the constant terms. We can find the
intersection by taking the inverse of the left matrix and multiplying on the left of both
sides of the equation. The simplified result gives
6
7
1
25 25 5
1
3 1
25 25 5
9
2
1
9
 2
2
2

1
3
25 25 5
This seems to be the most commonly taught method, and the one that students and
teachers seem to prefer, and yet it has two major disadvantages. The first disadvantage
is that it tells you little or nothing about systems which have a solution, but not a single
unique solution. The second is that the inverse method is more computationally complex,
that is, it takes more operations for the solution than the alternative RREF method, and
the difference grows as problems reach higher orders of magnitude. For the problems
that are generally assigned at the high school level, the difference in computability
presents no real problem, but the difference in the range of applicable questions can be
very significant in a students understanding of general systems of three equations.
The first obvious case is three planes intersecting in a single point that is the most
familiar to students. In addition to the Inverse method shown above, a second matrix
method using the augmented matrix to represent the system will also solve this system.
This method takes a matrix that combines the coefficient matrix with the constant matrix
and then uses the same row operations that students learned when they solved systems
by elimination to produce an augmented matrix with the solution easily apparent. Here is
1 1
2
2 9
3 1
2
the augmented matrix for the system, 1 3 1 2 , and the RREF matrix of the same
1 0 0 2
0 1 0 1
system
0 0 1
3
.
In contrast with the Inverse method that will only work if the three planes intersect
in a single point, the RREF form will allow us to work with systems which do not even
have the same number of equations as unknowns. This is the type of situation created
when we try to find the line of intersection of two planes.
RREF for two planes
We will use the equations 2x + 3y – 3z = 14 and –3x + y + 10z = -32. When we
write an augmented matrix for the system of only two equations we get a 2x4 matrix,
2 3
3 14
3 1 10
32
.
1 0
3 10
When we reduce this to the RREF we get 0 1 1 2 . Now how do we interpret
this result? If we convert the RREF of the augmented matrix back to a system of
equations we get x - 3z = 10 and y + z = -2 .
We notice that both equations contain a z variable, it might occur to us to ask,
“What happens if we substitute different values in for z?”. For example, if we try z=0 we
note that from the first equation we get x=10 and from the second we get y=-2. What
does this tell us about the point (10, -2, 0). If we check it against the two original
equations we notice that the point makes both equations true, 2(10) + 3(-2) – 3(0) = 14
and –3(10) + (-2) + 10(0) = -32. So the point (10, -2, 0) is on both planes and therefore
must lie on the line that is their intersection.
Can we find more points? What happens if we try z=1 or z=2 or other values.
Using z=1 we get x – 3(1) = 10 which simplifies to x=13 ; and y +(1) = -2 which
simplifies to y=-3. Checking the point (13, -3, 1) we see that it also makes both equations
true, and so it must also be on the line of intersection.
Writing the parametric equation of a line in three space
So now we have two points on the line of intersection; (10,-2, 0) and (13, -3, 1).
How can we describe the line? One way is to write a formula for all the points so that
someone could find as many values of (x,y,z) as they wish. We do this with parametric
equations. A parametric equation is an equation that explains the values of one set of
variables (in this case x, y, and z) in terms of another “parameter” which we will call t.
We will define the line in terms of one point, and instructions to get from one point to
another; sort of a three-space equivalent of slope.
When we worked with slope in a plane we had to find the change in x and the
change in y, but now in three space we need a change in z also. If we look at our two
known points, we can see that from the first we found, (10,-2, 0), to the second, (13, -3,
1), the x-value increases 3, the y-value decreases 1, and the z value increased 1. Let’s
record those as a ordered triple, but to keep it separate from our points we will use
brackets to enclose the changes; like this [3, -1, 1] .
We can even use this to find more points. If we add [3, -1, 1] to the last point we
found (13, -3, 1) we get (16, -4, 2); and we know it is on both planes because it makes
both the original equations true (Remember? They were 2x + 3y – 3z = 14 and –3x + y +
10z = -32) . Of course we do not have to add integer multiples of t, and if we allow t to be
ANY real value, then we can write a general expression for all the points on the line of
intersection in the form (x, y, z) = (10, -2, 0) + t [3, -1, 1] . [some texts will write this as
a vector form (x, y, z) = (10+3t, -2 – t, t ) ].
If we return to examine the RREF form of the augmented matrix we wonder if the
1 0
3 10
direction vector [3, -1, 1] is in anyway obvious from the reduced matrix 0 1 1 2 . We
are tantalized by the appearance of –3 and 1 in the Z column, and the 10, -2 in the
constants column. Could we have looked at this reduced form and gone directly to the
equation of the line? Perhaps a second example will help.
For the second two planes we pick { x+y + 5z = 8 and 2x + y + 8z = 13 }. Our
1 0 3 5
reduced row echelon form looks like 0 1 2 3 . Is it possible that our line of intersection is
as simple as (x, y, z) = (5,3,0) +t [-3, -2, 1] (clever folks are already thinking we could
write this last as [3, 2, -1] since t goes through all negative and positive values). Will the
points generated lie on both planes. (5, 3, 0) is almost a trivial to check, and if we check
(2, 1, 1) we see that it also is on both planes. We are now more assured that it is at least
plausible that the RREF matrix makes the intersection line simple to obtain.
A second consistent solution to three planes
Now let’s add to our system and make a system of three equations; {2x + 3y – 3z
= 14 , –3x + y + 10z = -32, x + 7y + 4z = -4}. If we try to solve this system by the
inverse method we get the dreaded “Singular Mat” error on the calculator and all we can
determine is that there is NOT a single unique solution. If we use the RREF we get
something that should remind us of our previous example, and for good reason. Because
I chose the third line to be a linear combination of the first two, the three planes all share
a common line of intersection, the line (x, y, z) = (10, -2, 0) + t [3, -1, 1] as shown by the
reduced form:
1 0
3 10
0 1 1
2
0 0 0
0
The third row gives us an obviously true statement that reflects that the third equation,
in essence, added no new information to the system. This “row of zeros” at the bottom is
the key to the student that the system had three equations but the intersection would be
completely determined by any two of them.
When three planes intersect in space, there are five basic configurations that can
take place. I ask students to visualize and describe the five situations that can exist prior
to discussion of the method of finding the solutions. In most years they come up with all
five configurations and names for them. Here is a list of the five as my students tend to
describe them.
The two consistent systems are three planes intersecting in a single point, and
three planes intersecting in a single line (the book binding model – the planes are pages in
a book and the common line is the binding).
The three configurations with no common points of intersection are three parallel
planes, two parallel planes cut by a transverse plane (the H configuration) forming two
parallel lines, and three planes intersecting in mutual pairs to form three parallel lines (the
delta  configuration). The first and the second are often easy to recognize from the
equations since the parallels are usually recognizable, but the third is difficult to identify
without reducing.
Three configurations w/o a common solution
If we look at the three inconsistent systems possible with three planes, the more
obvious cases may lead us to more understanding of the less obvious cases. First I want
to look at three parallel planes, x + 2y + 3z = 1, x+ 2y + 3z = 5, and x + 2y + 3z = 10.
The reduced RREF augmented matrix looks like this:
1 2 3 0
0 0 0 1
0 0 0 0
The first row looks like it is directly related to the coefficients of x, y, and z in each
equation, but the 0 in the constant term may seem perplexing. I think of this information
as a direction vector [1,2,3] that is perpendicular to the three parallel planes. The bottom
line of zeros reminds us that the last equation added no new information about the
solution set. But what are we to make of the equation for the second row, 0x + 0y + 0z
= 1. I tell students that this is the signal that there is NO common point of intersection.
In essence, we have an intersection when 0=1, and that is never true.
Now we examine a system with two of the same parallels and a third plane not
parallel to them, the H configuration, { x + 2y + 3z = 1, x+ 2y + 3z = 5, 2x – y + z = 1}.
1 0 1 0
0 1 1 0
The reduced matrix looks like 0 0 0 1 . How can we interpret this. The bottom row is
our old insolvable friend 0x +0y + 0z = 1, reminding us that there can be no common
point on the three planes (since two of them are parallel). The upper two rows look like
the rows when we had a line of intersection except that both constant terms equal zero.
Does this tell us anything about the two lines of intersection formed by the plane cutting
through the two parallel planes. Perhaps we can find out more if we find the two
intersections one at a time. If we reduce the two equations x + 2y + 3z = 1, 2x – y + z =
1 0 1 .6
1, our reduced matrix looks like this, 0 1 1 .2 . This gives a line (x, y, z) = (.6, .2, 0) + t
[-1, -1, 1] . If we do the intersection with the second parallel plane we get (x, y, z) =
(1.4, 1.8, 0) + t [-1, -1, 1]. It seems that our original reduced system was flashing the
direction vector part of the two parallel lines created by the three planes in the third
column.
For our final case, we pick the inconsistent situation which is most difficult to
distinguish from the consistent cases, the delta  configuration in which three planes
intersect in mutual pairs to form three parallel lines. An example is the three planes
defined by { x + y + 7z = 5, x – y – z = 3, and 2x – y + 2z = 4} .
1 0 3 0
0 1 4 0
The reduced form looks like this, 0 0 0 1 and we are now becoming accustom to
the 0001 on the bottom reminding us there is no common point for the three planes. And
a quick inspection of the three planes assures us that no two of them are parallel, so the
only possible case is the “delta formation” and three parallel lines. What are we to make
of the two upper rows. Our experience now leads us to think that [3,4,-1] will be the
direction vector for all three lines. We can verify our supposition by taking the vectors in
pairs and reducing the two-equation systems to find the actual line of intersection, but as
is the fashion in mathematical writing, I leave that final task as an exercise for the reader.