# How to determine if a linear system is solvable

I have this problem $$\begin{array}{rcccccl} 3x &-&y &+& 2z &=& 2\\ 2x &+& y &+& z &=& -1\\ x &+& 3y & & &=&-1 \end{array}$$ This gives me the matrix: $$\left(\begin{array}{rrr|r} 3 & -1 & 2 & 2\\ 2 & 1 & 1 & -1\\ 1 & 3 & 0 & -1 \end{array}\right)$$

I remember something about if you can show that a system in inconsistent, you know it's not solvable.

I set Row1->Row1 - 2*Row2 + Row3

The result is: $$\left(\begin{array}{rrr|r} 0 & 0 & 0 & 3\\ 2 & 1 & 1 & -1\\ 1 & 3 & 0 & -1 \end{array}\right)$$

So 0 = 3 which is inconsistent.

Two things, did I do this right and is my assumption that all linear systems that can be shown as inconsistent are unsolvable?

Are there less painful ways to show it's not solvable? It seems like this could go on for a while with trial and error to either find an inconsistency or an answer for x y and z.

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Row reduction is a process which can easily be encoded in an algorithm. And once you get the so called "Recuced Row Echeleon Form", that tells you if the system is consistent or not, and moreover how to get the generals solution... – N. S. Feb 2 '12 at 3:43
The matrix isn't correct. The last row must be | 1 3 0 | -1 | – leo Feb 2 '12 at 3:53
@leo fixed that. it was a typo – brandon Feb 2 '12 at 4:07
@N.S. I know I can see if the system is inconsistent this way. My question is more what does it mean if it's inconsistent. Does it mean it doesn't have a unique solution? – brandon Feb 2 '12 at 4:08
@brandon: You had not fixed the typos; but your operations were only correct had the matrix been the correct one, so I fixed it. – Arturo Magidin Feb 2 '12 at 4:31

1. Yes: by showing that the system is equivalent to one in which the equation $0=3$ must hold, you have shown the original system has no solutions.

2. By definition, a system of linear equation is said to be "consistent" if and only if it has at least one solution; and it is "inconsistent" if and only if it has no solutions. So "showing a system of linear equations is not solvable" (has no solutions) is, by definition, the same thing as showing that the system of linear equations is "inconsistent".

3. "A system doesn't have a unique solution" can happen in two ways: it can have more than one solution (in which case it has infinitely many solutions), or it can have no solutions. Only in the second case do we say the system is "inconsistent".

4. One of the easiest ways to find solutions of systems of linear equations (or show no solutions exist) is Gauss (or Gauss-Jordan) Row Reduction; it amounts to doing the kind of things you did, but in a systematic, algorithmic, recipe-like manner. You can do it almost mindlessly (or you can keep an eye out for clever actions that may save you work down the line). This process will determine, after a finite number of steps, whether the system is consistent or not, and in the case that the system is consistent, will tell you what the solutions are.

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The idea of the row operations is that the solutions, if any, after the operations are exactly the same as the solutions, if any, before the operations. If, after the operations, you get to an equation like $0=3$, that is, to an equation that obviously has no solutions, then you deduce that your original system had no solutions. And if your original system has no solutions, then a systematic application of row operations will pretty quickly get you to an equation like $0=3$.

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If the determinant of the left hand matrix is 0, then the system is dependent and has a free variable. This means the system has an infinite number of solutions.Finding if a system has no solutions is rather more complex. If one can write one row of the left hand side of the matrix as a combination of the other rows and (after doing so) one has a different value on the right hand side of the linear combination than he/she does from the original, there are no solutions.

I'm not sure how to add a matrix here to demonstrate the concepts.

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Well, the solution is unique when there are no free variables. Given an augmented matrix $M$, transform $M$ to row echelon form. If you have an entire row of zeros to the left of the vertical bar and a non-zero entry to the right of the bar, then the system is inconsistent.

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Rank based decision is a straight forward approach and will work always. If rank(A) = rank(A|b), then a solution exists.

How to Check for Existence of Solution to Matrix Equations

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