Linear Algebra – How to Determine if a Linear System is Solvable

linear algebra

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.

Best Answer

  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.

Related Question