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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 $0$ ?

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    $\begingroup$ By cursory inspection, adding the first two, then the last two equalities, both give the same equation $\,a+b+c+d=34\,$, so the system is redundant. $\endgroup$ – dxiv Jan 30 '18 at 6:48
  • $\begingroup$ For there to be precisely 4 solutions the equations must be linearly independent. These ones aren't. As dxiv's comment points out you can combine two to get a combination of the other two. $\endgroup$ – fleablood Jan 30 '18 at 7:01
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    $\begingroup$ @dxiv Thank you so much, you are right. $\endgroup$ – Dima Jan 30 '18 at 7:29
  • $\begingroup$ @fleablood thank you so much, your comment is helpful. $\endgroup$ – Dima Jan 30 '18 at 8:04
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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.

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  • $\begingroup$ Thank you so so so much, Your answer is so helpful. $\endgroup$ – Dima Jan 30 '18 at 8:03
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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.

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  • $\begingroup$ Thank you so much, you are right 👌🏻 $\endgroup$ – Dima Jan 30 '18 at 8:04
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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 !).

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  • $\begingroup$ Thank you, That’s what I did, but I cannot make it Echolen form :(, maybe because the equations not independent ? $\endgroup$ – Dima Jan 30 '18 at 7:28
  • $\begingroup$ Then proceed step by step from left to right ... $\endgroup$ – Duchamp Gérard H. E. Jan 31 '18 at 9:55

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