Number of Equations Needed in a Simultaneous Linear System

Date: 10/29/2003 at 08:37:55
From: Godfried
Subject: TRICK question about systems of linear equations

Hi,

Could you tell me why we need the same number of equations as 
variables in order to get a unique solution to a system of 
simultaneous linear equations? 
 
Some people say it's simply a fact...but I'd like to know why.

Date: 10/29/2003 at 11:43:53
From: Doctor Douglas
Subject: Re: TRICK question about systems of linear equations

Hi Godfried.

Thanks for writing to the Math Forum.  Understanding why things are
true is certainly better than just accepting them as fact.

First let's look at one variable:  x.  If you had one linear equation 
involving x, you'd know what it is:

  3x = 12        ->      clearly x = 4

If you had another equation in that system, such as -2x = 10, this
would not give a consistent solution for x.  The variable x cannot
simultaneously equal 4 and -5.

Now let's consider a system of two variables:  x,y.  Two equations are 
necessary to come to a unique solution:
   
   x + y = 6
  2x - y = 0

In this case, you can easily solve the two equations in two unknowns
for x and y, and find that x = 2, y = 4.

In general, more equations added to this system will make the solution 
disappear, as the system becomes "overconstrained":

   x + y = 6   solving the first two equations together implies x = 2.
  2x - y = 0   solving the last two equations together implies x = -1.
   x - y = 1

Clearly, the added equation does not help.  There are too many 
constraints on the variables x and y.

On the other hand, fewer equations don't provide enough constraints
for a unique solution.  If we have only the first equation alone 
(i.e., x + y = 6), this tells us something about the relationship
between x and y, and allows x = 2 (and y = 4) as a possible solution, 
but it is not a unique solution.  In fact, there would be infinite
possible solutions to that equation by itself.

A similar type of reasoning applies as we increase the number of
variables to three, four, and so on.  We will need one additional
equation for each additional variable.
 
Remark I:  it is not always true that more equations than variables
  leads to no solution.  Sometimes, we can get lucky:

     x + y = 6
    2x - y = 0
    4x - y = 4  this added equation is "consistent" with x = 2, y = 4.

  However, this situation does not usually happen.

Remark II:  it is not always true that having exactly N linear
  equations for N variables always leads to a unique solution.  It
  is possible to have either no solution (inconsistent) or an
  infinite number of solutions (underconstrained):

  Examples with N = 3:

     x + y + z = 8                 x + y + z = 8
     x + y     = 4              2x + 2y + 2z = 16
     2x + 2y   = 5                         z = 5

     This system is             This system is underconstrained.
     inconsistent.              We know the exact value of only z.
            
I hope this helps.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/30/2003 at 05:05:24
From: Godfried
Subject: TRICK question about systems of linear equations

Thanks for your answer; you clarified a lot.  I also found a geometric
explanation, here it is:

"If you have the  graphs of 2 linear equations with 2 variables (x,y), 
there's almost always an intersection point, which is the unique 
simultaneous solution."

\  /
 \/
 /\
/  \

Also, if you have equations in 3D space, you need 3 equations instead 
of 2.  But, if 2 lines can intersect in 3D space (which I think is 
possible), what's the need, then, for a third line?

Date: 10/30/2003 at 10:10:05
From: Doctor Douglas
Subject: Re: TRICK question about systems of linear equations

Hi again, Godfried.

Excellent followup!  In 3D, the relevant geometric objects are planes,
not lines:

  Ax + By + Cz = D       <----     a plane

and it is not too hard to see that 2 planes that intersect in 3D space 
do so in a (1D) line, and that a third plane, if it intersects that 
line, will generate the point that is the unique solution to all three 
linear equations. 

In 4D, the relevant geometric objects are 3D-spaces.  They mutually 
intersect in 2D-planes.  Three of these spaces will in general specify 
a 1D-line and four of these spaces will in general determine the point 
that is the unique solution to all four linear equations.

Tying the geometric interpretation into some ideas from my first 
reply, note that parallel lines in 2D relate to an inconsistent 
system:

  y = 2x + 3
  y = 2x - 4

Similarly, two lines which are identical relate to an underconstrained
system since a unique intersection point cannot be determined.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/30/2003 at 11:19:04
From: Godfried
Subject: Thank you (TRICK question about systems of linear equations)

Thanks for your clear explanation, and for preventing me drawing any
more lines in 3D space. ;)