Showing a linear system has no solution

Question

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?'.

Answer 1

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 $10$.

Answer 2

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 $x_1, x_2, x_3$).

Answer 3

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.