System of $4$ equations

Question

first I don’t speak English well, so sorry for that.

I have a question about this system:

$a+b=22$

$c+d=12$

$a+c=14$

$b+d=20$

This system has $4$ equations and $4$ variables, so I said it has unique solution, but I found More than one solution, why ? Is this because coefficients of $a,b, c, d$ in some equations are equal to $0$ ?

Answer 1

The fallacy here is thinking that having the same number of equations as variables guarantees that you will have a unique solution. However, it's possible to have no solution or to have infinitely many solutions. Here is an example of each for only 2 equations and 2 variables:

1) $x+y=1,x+y=2$ has no solutions.

2) $x+y=1,x+y=1$ has infinitely many solutions.

Having at least as many equations as variables is a necessary condition for a unique solution, but it is not sufficient. In particular, if we can take each of the equations, multiply them each by some number (not all 0), and add them all up to get a constant equation $a=b$, then there won't be a unique solution. If $a$ and $b$ are actually equal (ex: $3=3$), then the solution exists but is not unique. If $a$ and $b$ are distinct (ex: $2=9$), then there will be no solution. If there is no way to get such an equation by this process, then there will be a unique solution.

This process is called a linear combination, and the equations are called linearly dependent when some linear combination creates a constant equation, and linearly independent when no linear combination does.

Answer 2

You can use standard matrix calculations to find the rank of your system which amounts to $$ \begin{pmatrix} 1 & 1 & 0 & 0\cr 0 &0&1&1\cr 1&0&1&0\cr 0&1&0&1 \end{pmatrix} \begin{pmatrix} a\cr b\cr c\cr d \end{pmatrix} = \begin{pmatrix} 22\cr 12\cr 14\cr 20 \end{pmatrix} $$ Hint By the above you see that (a) there is a solution (b) the kernel is of dimension one (use Gaussian elimination !).

Answer 3

It is not enough for a system to have as many equation as there are unknowns for the solution to be unique. The equations must also be independent, otherwise there is too little information.

If you add the first two equations and subtract the third, you obtain

$$b+d=20$$ which is exactly the fourth. So what you have is actually equivalent to a system of three equations in four unknowns.