Systems of linear equations

I am I right on the following:

A 2X2 system of equation with no solution will look like this:

x+3y=3
2x+6y =-8
1/2 + 3/6 ≠ 3/8


A 2X2 system of equations with in finite solutions will look like this:

x +2y=4
2x+4y=8
1/2 + 2/4 =4/8


A 2X2 system of equations with unique solution will look like this

x  -y= 1
3x +2y=12


This is because they are multiples of each other?

Thanks!

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Why do you say 4x4 when all your examples are 2x2? – Ted May 13 '12 at 19:03
Oops, i should change that. 10 hours of just linear algebra can really get to your head :/ – Xabi May 13 '12 at 19:29
How do you get $1/2 + 3/6 \neq 3/8$ and $1/2 + 2/4 =4/8$? And the latter is wrong. – TMM May 13 '12 at 19:33

An $m\times n$ system of $m$ linear equations in $n$ unknowns has no solutions when the system can be transformed into an equivalent system in which one of the equations is false. In this case, the system is called inconsistent. A system is inconsistent iff there is no solution.

A system has infinitely many solutions when, after simplification to remove redundancies, for example with row reduction of the equivalent matrix, you are left with less equations than unknowns. This is sometimes called underdetermined (such systems have free variables, which can be used to parametrize the solution space).

When you are left with the same number of reduced (or independent) equations as unknowns, the system has a unique solution.

The process of reducing a system of equations to eliminate redundancies can be carried out systematically, in fact with matrices, using columns to represent each variable or the constant on the RHS. The corresponding process with matrices is called row reduction or Gaussian elimination.

The $2\times2$ system in your first example $$\matrix{ x&+&3y&=&3\\ 2x&+&6y&=&-8\\ }$$ can be simplified/reduced by subtracting a multiple of the first equation from the second, to get the equivalent system: $$\matrix{ x&+&3y&=&3\\ 0x&+&0y&=&-14\\ }$$ Since the second equation $0=-14$ is false, the system has no solutions. Note that if you divided the two original equations, you would get an equivalent falsehood (all falsehoods are equivalent): $$\frac12=\frac{x+3y}{2(x+3y)}=\frac{x+3y}{2x+6y}=\frac3{-8}=-\frac38,$$ however you cannot distribute the division over each term!

If your second example were $$\matrix{x&+&2y&=&4\\2x&+&4y&=&8,}$$ this would be a good example of an underdetermined, but consistent system with infinitely many solutions, since the second equations is a multiple of the first, therefore redundant (contains the same information). The solutions can be parametrized as $x=4-2y$ for $y\in\mathbb{R}$ free. Or you could say $x=4-2t,~y=t$ for $t\in\mathbb{R}$ free. Again, here, you cannot divide the like parts of the first two equations to obtain another equivalent equation.

Your third example is fine. But these are all $2\times2$ examples -- two equations, and two unknowns.

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Overdetermined is the wrong word. A system with more equations than unknowns is called overdetermined, even if it does have solutions. What you want is "inconsistent". And to clarify slightly: the other two cases require first of all that the system is not inconsistent. It doesn't matter how many equations and unknowns you have if the equations are inconsistent. – Robert Israel May 13 '12 at 18:14
@RobertIsrael: Thanks. – bgins May 13 '12 at 18:37