$$ \begin{array}{lcl} 2x + 2y + z&=& 0 \\ 2x − 3y − 4z&=& 0 \\ 4x − y − 3z &=& 0 \end{array}$$
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)?
$\endgroup$ 32 Answers
$\begingroup$The rank of a homogeneous system is the number of nonzero rows in its reduced row-echelon form.
Fact. If the rank of a homogeneous system is less than the number of variables in the system, then the system has infinitely many solutions.
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.
$\endgroup$ $\begingroup$This system of equations can be written as $AX=0$ where $A=\begin{pmatrix} 2&2&1\\2&-3&-4\\4&-1&-3 \end{pmatrix}$ and $X=\begin{pmatrix} x\\y\\z\end{pmatrix}$. Notice that $X=0$ is always a solutions of such a system, hence homogeneous systems always have at least one solution. Now if $A$ is invertible, then $AX=0\Rightarrow X=A^{-1}0=0$, hence this system has the unique solution $X=0$. If $A$ is not invertible, the system has more infinitely many solutions (this is not so difficult to prove, but requires some basic knowledge of inverses). So you can easily check whether such a system has a unique solution or infinitely many solutions by calculating $\det(A)$.
Remember this trick, if you're not explicitly interested in the solution, but rather in the number of solutions, then the determinant is very handy.
Also, as remarked in the comments, the equations need not to be scalar multiples of each other to carry the same information. For example: $x+y=0, x+y+z=0, z=0$, the third equation is redundant, since by the second we have that $(x+y)+z=0$ and by the first $x+y=0$, hence $z=(x+y)+z=0$ yields the third.
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