1.6 Constraints vs Equations

This should be clear by now, but it is still good to note that this categorization against the number of variables vs the number of equations is not corollated to being consistent or inconsistent systems. In fact, they very much overlap as we have seen. What matters, generally, is the number of constraints against the number of variables.

We are still interested in the number of equations because that is what we see first. Let us look at simple examples and see how consistency overlaps with how determined the system is. (Some of the following will serve as a summary)

1.6.1 Underdetermined and Inconsistent

\[ x + y + z = 5\] \[ x + y + z = 10\]

It should be intuitive that this is inconsistent. Both equations have the same LHS but different RHS. One thing cannot equal unequal things. Solving this system would lead to \(5 = 10\) which is contradictory so, this system is inconsistent.

By definition, an underdetermined system has fewer equations than variables, and consequently, fewer constraints than variables. So, the only way for an underdetermined system to be inconsistent is to have contradicting constraints.

1.6.2 Underdetermined and Consistent

\[ x + y + z = 5\] \[ x + y + 2z = 10\]

This looks similar to the last system except the second equation has \(2z\), which removed the contradiction. This now has an infinite number of solutions as there are fewer constraints than variables. To intuitively solve for this, we could rewrite it as:

\[ x + y + z = 5\] \[ (x + y + z) + z = 10\]

Where what is inside the parenthesis is the same as the first equation. So, we can replace it as such:

\[ x + y + z = 5\] \[ (5) + z = 10\]

So \(z=5\). If we plug \(z\) back in and simplify, we get \(x+y=0\). Then, we have infinite number of solutions where \(x\) and \(y\) add up to \(0\). Also, this equation is both undetermined and homogeneous so, it has the trivial solution and infinitely many others. (Recall Corollary (c) of Theorem 1.3 and Theorem 1.2).

Another example of a consistent underdetermined system is a single equation:

\[ x + y = 5\]

This is consistent because we have infinite solutions for any two numbers that add up to \(5\). Yet another example with 3 unknowns is:

\[ x + y + z = 5\]

Again, an underdetermined system always has fewer constraints than variables, and therefore, it is consistent as long as there are no contradictions.

1.6.3 Determined and Inconsistent

This happens again when we have contradicting constraints such as the following:

\[ x + y = 5\] \[ x + y = 6\]

1.6.4 Determined and Consistent

\[ x + y = 5\] \[ x + 2y = 6\]

This is consistent with a unique solution because the number of constraints equals the number of variables. We can solve for this by substitution or elimination to get \(x=4\) and \(y=1\).

Another example is one that has infinite number of solutions:

\[ x + y = 5\] \[ 2x + 2y = 10\]

That is because the two equations are equivalent, so the number of constraints is fewer than the number of variables.

1.6.5 Overderdetermined and Inconsistent

\[ x + y = 5\] \[ 2x + 2y = 5\] \[ 3x + 3y = 15\]

This is inconsistent as the second equation contradicts the other two.

\[ x + y = 5\] \[ 2x + y = 5\] \[ x + 3y = 15\]

This is inconsistent because we have more constraints than variables.

An overdetermined system is inconsistent when we have more constraints than variables or when there are contradicting constraints.

1.6.6 Overderdetermined and Consistent

\[ x + y = 5\] \[ 2x + 2y = 10\] \[ 3x + 3y = 15\]

All these equations here are equivalent; we can divide the second one by \(2\) and the third one by \(3\) to get the first one. So, we have 2 variables and 1 constraint. The system therefore is consistent with infinite number of solutions.

\[ x + y = 5\] \[ 2x + y = 10\] \[ 3x + 3y = 15\]

This now has two constraints but still two variables. So, this is consistent with a unique solution.

\[ x + y = 0\] \[ 2x + y = 0\] \[ x + 3y = 0\]

This is consistent because it is homogeneous even though it has more constraints than variables

In general, if the constraints are fewer than the variables, we have infinite number of solutions. If the number of constraints equal the number of variables, we have a unique solution. If the number of constraints is more than the number of variables, we have no solutions. But in any case, if we have contradicting constraints, then there is no solution. Also, as we know, a homogeneous system is always consistent.

We could state that here as a theorem with exceptions but we will cover a more general theorem later once we develop the tools needed. That will serve as a quick systematic/mechanical check later. This, however, equips us with understanding the building blocks of systems of equations.