*three* variables.

**not** equivalent to the two equation system?

In the last problem, we saw that given two equations in three unknowns, we can reduce to a single equation with two unknowns (and eliminate whichever of the variables we’d like). Thus, with $3$ equations with $3$ unknowns, we can reduce to two equations in the same two unknowns, which we already know how to handle!

Suppose $x, y, z$ satisfy $\begin{aligned} 3x + 2y + z &= 10\\ 4x - 3y + 4z &= 10\\ 5x - 4y + 2z &= 3. \end{aligned}$ What is $x + y + z?$

Of course, as before it is possible for this method to “fail,” in the sense that the system may no longer have a single solution. For instance, if we have the system $\begin{aligned} 3x + 2y + z &= 10\\ 6x + 4y + 2z &= 20\\ 5x - 4y + 2z &= 3. \end{aligned}$ Then there is no unique solution since the second equation is twice the first (similar to what we saw in the previous quiz). But with $3$ equations, it is not always so easy to detect when this happens, as we see below.

Suppose we have the system
$\begin{aligned}
3x + 2y + z &= 10\\
4x - 2y + 4z &= 10\\
5x + y + az &= 15.
\end{aligned}$
What value of $a$ would lead this system to **not** have a unique solution?

By now you’ve probably noticed an interesting pattern: to solve a system in two variables, we (usually) need two equations, and to solve a system in three variables, we (usually) need three equations. This leads to a natural question: do we need four equations to solve a system in four variables?

It turns out that the answer is yes: we need at least as many equations as there are **degrees of freedom** in the system. We’ll come back and formally define this later, but intuitively degrees of freedom correspond to the number of independent variables in the system. For example, if there are *no* equations in a $3$-variable system, obviously $x, y,$ and $z$ can be anything--so they’re all independent, and there are $3$ degrees of freedom. If there is one equation, then when we pick $x$ and $y,$ the value of $z$ is (usually) forced, so $z$ is no longer independent and there are only $2$ degrees of freedom left. To get rid of all the degrees of freedom — making a unique solution — we need $3$ equations.

In the previous quiz, we saw how $2$-variable equations can be represented by lines, and solving a system of equations can be viewed as finding the intersection of these lines. Similarly, a $3$-variable equation can be viewed as a plane, and solving a $3$-variable system can be viewed as finding the intersection of these planes.

This also shows why there are more “exceptions,” or degenerate systems, to the general rule of $3$ equations being enough for $3$ variables. Which of the following images of planes does **not** correspond to a degenerate system of equations?

A)

B)

C)

D)