FMCC Going Deeper: Three Dimensional Systems

Section 16.5 Going Deeper: Three Dimensional Systems

In the last couple sections, we've looked at systems of two linear equations with two variables. We saw that there was a geometric aspect to understanding these systems by looking at the graphs of the lines, and we also saw that there was an algebraic aspect by looking at the methods of solving these systems. We are going to take a look at what happens when we increase the complexity by looking at systems of three equations and three variables. Here is an example:

\begin{equation*} \left\{ \begin{array}{rcrcrcr} 2x \amp - \amp y \amp - \amp z \amp = \amp 1 \\ x \amp + \amp 2y \amp + \amp z \amp = \amp 4 \\ x \amp + \amp y \amp - \amp z \amp = \amp 4 \end{array} \right. \end{equation*}

We will first wrap our minds around the geometric understanding of these equations. It turns out that these formulas form planes and that planes are the natural shape to generalize the "linear-ness" of lines in two variables. When we think about three simultaneous systems of equations in three variables, we're looking for a point that lies on all three planes. Since the number of dimensions has increased, it makes sense that the number of possible arrangements would also increase. However, we still have the three same basic classifications that we had before. There will either be zero solutions, one solution, or infinitely many solutions.

In each of the systems above, there are no solutions. This may seem unusual at first, since there are clearly some points of intersection. However, we are working with simultaneous systems of equations, which means that we're looking for points that lie on all three planes, not just pairs of planes. Notice that we cannot use the simple heuristic of just thinking about whether the planes are parallel, because the diagram on the right shows that it's possible to have no solutions even though none of the planes are parallel to each other.

A key observation about these diagrams is that when there are intersections between planes, it forms a line. This is always true of the intersection of two planes, which is an idea we'll come back to when we look at the algebra.

This is an example of a system with one solution. Although everything is drawn to be at right angles with each other, there is a lot of flexibility in terms of the relative positions of the planes. You should be able to visualize each pair of planes forming a line, and that each of those lines cross each other in exactly one place.

Lastly, we have these configurations to get infinitely many points of intersection. The diagram on the left has all three planes overlapping each other. The middle diagram has two planes overlapping each other with the third plane cutting across them. The last diagram has all three planes intersecting along the same line. Even though these all have infinitely many solutions, there is a distinction between the first one and the last two. Intuitively, what we're seeing are different numbers of dimensions in the overlap. The intersection of the first diagram is a two-dimensional plane, whereas the intersection of the last two are a one-dimensional line.

When it comes to solving these systems algebraically, it's just the methodical application of either substitution or elimination to reduce the number of variables. Conceptually, we're going to find the intersection of planes to get lines, then find the intersection of lines to get points. When working through the process, it's much more about keeping the work organized so that you don't get lost.

Subsection 16.5.1 Solving Three Dimensional Systems by Substitution

We will solve the first equation for \(z\text{:}\)

\begin{equation*} \begin{aligned} 2x - y - z \amp = 1 \\ z \amp = 2x - y - 1 \end{aligned} \end{equation*}

Then we will substitute this into the other equations and simplify:

\begin{equation*} \begin{aligned} x + 2y + z \amp = 4 \\ x + 2y + (2x - y - 1) \amp = 4 \\ 3x + y = \amp 5 \end{aligned} \qquad\qquad \begin{aligned} x + y - z \amp = 4 \\ x + y - (2x - y - 1) \amp = 4 \\ -x + 2y \amp = 3 \end{aligned} \end{equation*}

We can now solve for \(y\) in the first of the new equations:

\begin{equation*} \begin{aligned} 3x + y \amp = 5 \\ y \amp = -3x + 5 \end{aligned} \end{equation*}

And plug that into the second of the new equations and solve for \(x\text{:}\)

\begin{equation*} \begin{aligned} -x + 2y \amp = 3 \\ -x + 2(-3x + 5) \amp = 3 \\ -7x + 10 \amp = 3 \\ x \amp = 1 \end{aligned} \end{equation*}

We can now work backwards to get the remaining variables.

\begin{equation*} \begin{aligned} y \amp = -3x + 5 \\ \amp = -3(1) + 5 \\ \amp = 2 \end{aligned} \qquad\qquad \begin{aligned} z \amp = 2x - y - 1 \\ \amp = 2(1) - (2) - 1 \\ \amp = -1 \end{aligned} \end{equation*}

The solution of the system of equations is \((x,y,z) = (1,2,-1)\text{.}\)

Subsection 16.5.2 Solving Three Dimensional Systems by Elimination

We will first eliminate \(z\) twice by first combining the first two equations then the last two equations:

We will now combine these equations to eliminate \(x\text{:}\)

Then combine them again to eliminate \(y\text{:}\)

While we could be stubborn and go through this process again to solve for \(z\) using elimination, it makes much more sense to simply substitute the known values into one of the original equations to solve for \(z\text{.}\) We will use the first equation.

\begin{equation*} \begin{aligned} 2x - y - z \amp = 1 \\ 2(1) - (2) - z \amp = 1 \\ -z \amp = 1 \\ z \amp = -1 \end{aligned} \end{equation*}

The solution of the system of equations is \((x,y,z) = (1,2,-1)\text{.}\)