The number of solutions of the equations
$x_{2}-x_{3}=1$
$-x_{1}+2x_{3}=2$
$x_{1}-2 x_{2}=3$
(a) zero (b) one (c) two (d) infinite
In a solution,I saw $$ \begin{array}{l} \text { Let } \Delta=\left|\begin{array}{ccc} 0 & 1 & -1 \\ -1 & 0 & 2 \\ 1 & -2 & 0 \end{array}\right| \\ =-1(-2)-1(2)=0 \end{array} $$ Hence given system of equations is consistent and since have no two equations are identical, therefore number of solutions is one. So $\boxed{Ans. (b)}$
My approach
$\left[\begin{array}{ccccc}0 & +1 & -1 & 1 \\ -1 & 0 & 2 & 2 \\ 1 & -2 & 0 & 3\end{array}\right]$
$\left[\begin{array}{ccccc}0 & 1 & -1 & . & 1 \\ -1 & 0 & 2 & . & 2 \\ 0 & -2 & 2 & . & 5\end{array}\right] \quad \quad R_{3}=R_{2}+R_{3}$
$\left[\begin{array}{cccc}0 & 1 & -1 & 1 \\ -1 & 0 & 2 & 2 \\ 0 & 0 & 0 & 7\end{array}\right] \quad R_{3}=2 R_{1}+R_{3}$
$\rho(A) \neq \rho(A \mid B)$
So NO Solution
I am getting a contradiction. Am I wrong