$$
\begin{array}{lcl}
2x + 2y + z&=& 0 \\
2x − 3y − 4z&=& 0 \\
4x − y − 3z &=& 0
\end{array}$$
What I've gathered thus far from my book is that a system of linear equations has infinitely many solutions if (this is just how I've worked it out in my brain; this isn't actually written in my book or anything and could be wrong):
1) The system has more variables than it has equations
or
2) All the equations in a system are scalar multiples of each other
But neither seems to me to be the case here, since…
There are 3 variables, as well as 3 equations.
And 2x+2y+z=0, for example, isn't a scalar multiple of 4x-y-3z = 0.
So why does this system they have infinitely many solutions, and how can I recognize this (if I even need to)?
Best Answer
The rank of a homogeneous system is the number of nonzero rows in its reduced row-echelon form.
In our example, we have $$ \DeclareMathOperator{rref}{rref}\rref \left[\begin{array}{rrr} 2 & 2 & 1 \\ 2 & -3 & -4 \\ 4 & -1 & -3 \end{array}\right]= \left[\begin{array}{rrr} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right] $$ The rank in this case is two, while the number of variables is three. Hence the system has infinitely many solutions.
In fact, the reduced row-echelon form can be used to explicitly describe these infinitely many solutions. Here, all solutions are of the form $$ \left[\begin{array}{r} x \\ y \\ z \end{array}\right]= \left[\begin{array}{r} z/2 \\ -z \\ z \end{array}\right] = z \left[\begin{array}{r} 1/2\\-1\\1\end{array}\right] $$ Evidently, the dimension of the solution space is one. In general, the solution space has dimension equal to the number of variables minus the rank.