solve a linear system using gauss-jordan elimination method We are asked to solve the following linear system
$$x_1-3x_2+x_3=1$$
$$2x_1-x_2-2x_3=2$$
$$x_1+2x_2-3x_3=-1$$
by using gauss-jordan elimination method. The augmented matrix of the linear system is $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\2 & -1 & -2 & 2 \\1 & 2 & -3 & -1\end{array}\right).$$ By a series of elementary row operations, we have $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right).$$ My question is, although the question asked us to solve the linear system using gauss-jordan elimination method, can we stop immediately and conclude that the linear system is inconsistent without further apply any elementary row operation to the matrix $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right)$$ until the matrix $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right)$$ is transformed into reduced-row echelon form?
 A: You can conclude that the system is inconsistent, because the last row of your matrix implies that $0x_1+0x_2+0x_3=-2$, which cannot be satisfied.
A: Yes, you can stop there and conclude that the system is inconsistent as $0\ne-2$. If you were to continue to reduce the matrix to reduced-row echelon form, row $3$'s inconsistency would remain unaffected.
$$\left(\begin{array}{ccc|c}1 & 0 & -\frac{7}{5} & 0 \\0 & 1 & -\frac{4}{5} & 0 \\0 & 0 & 0 & 1\end{array}\right)$$
$R_3\to-\frac{1}{2}R_3$ was performed to get the new row $3$ and notice that the completely reduced-row echelon form above also has $0x_1+0x_2+0x_3=1 \implies 0=1$ which is not possible, and hence the system still maintains its inconsistency. 
A: $$ rank \left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\2 & -1 & -2 & 2 \\1 & 2 & -3 & -1\end{array}\right)=3$$ because $$det\left(\begin{array}{ccc}-3 & 1 & 1 \\-1 & -2 & 2 \\2 & -3 & -1\end{array}\right)\neq 0$$ and the rank of the coefficient matrix $$\left(\begin{array}{ccc}1 & -3 & 1 \\2 & -1 & -2 \\1 & 2 & -3 \end{array}\right)$$ is $2$ because its determinant is $0$ and $$det\left(\begin{array}{cc}1 & -3 \\2 & -1 \end{array}\right)\neq 0$$ 
Augmented matrix and coefficient matrix have different ranks then by the Rouchè-Capelli theorem the system has no solutions
