# Showing a linear system has no solution

The system in question is $$\begin{cases} x_1 -x_2 + x_3 = -1 \\ -3x_1 +5x_2 + 3x_3 = 7 \\ 2x_1 -x_2 + 5x_3 = 4 \end{cases}$$

After writing this in matrix-form and performing row-operations we can show that

$$\begin{matrix} -1 & 4 & 8 &| 11 \\ 0 & 1 & 3 &|6\\ 0 & 1 & 3&|10 \\ \end{matrix}$$

Substitute back our variables and we get

$$\begin{cases} x_2 + 3x_3 = 6 \\ x_2 +3x_3 = 10 \end{cases}$$

Which is a contradictory statement that shows our system has no solutions. Is this a suffciently rigorous way of answering the question : 'Does this system have a solution?'.

• The contradiction should be that $10\neq 6$. However, in $\Bbb F_2$ this is no contradiction. You need to mention over which field you are working. – Dietrich Burde Nov 6 at 10:36

Yes, that is sufficient, since at that point you can assert that, no matter what the values of $$x_2$$, and $$x_3$$ are, the number $$x_2+3x_3$$ cannot be equal to both $$6$$ and $$10$$.

• Yes, this is true. It would be nice, though, if the askers were aware that there are different fields. – Dietrich Burde Nov 6 at 10:43

You can do one step more and transform the last line to $$(0, 0, 0 | 4)$$ so that it's clear that the system is not compatible and hence admits no solution (contradiction $$0=4$$ for any value of the variables $$x_1, x_2, x_3$$).

• For coding theory, over the finite field $\Bbb F_2$, your conclusion is not true. – Dietrich Burde Nov 6 at 10:38
• It's not stated explicitly in the question, but I believe the OP is in $\mathbb R$. – G. Gare Nov 6 at 10:41
• Homework regarding gaussian elimination for a Linear Algebra course – juhani Nov 6 at 10:43
• @juhani For Linear Algebra, what fields do you know? – Dietrich Burde Nov 6 at 10:44
• .. and that was quite clear from the beginning. @DietrichBurde, if my little cousin comes to me and asks me what is 10 times 10 I don't ask her to specify whether she's in basis 10 or in binary – G. Gare Nov 6 at 10:44

Using the Kronecker-Capelli criterion, the system of equations is solvable if and only if the rank of the system matrix is equal to the rank of the augmented matrix. The rank of the system matrix is clearly $$2$$ and of the augmented matrix is $$3$$. Therefore, there can be no solution.