Answer
See below
Work Step by Step
$\begin{bmatrix}
1 &1 &1|y_1\\
2 &3 &1|y_2\\
3&5&1|y_3
\end{bmatrix}\approx \begin{bmatrix}
1 &1 &1|y_1\\
0 &1 &-1|y_2-2y_1\\
0 & 2& -2|y_3-3y_1
\end{bmatrix} \approx \begin{bmatrix}
1 &1 &1|y_1\\
0 &1 &-1|y_2-2y_1\\
0 & 0& 0|y_3-2y_2+y_3
\end{bmatrix}$
$1.A_{12}(-2),A_{13}(-3)\\
2.A_{23}(-2)$
We must have $rank A=rank (A*)$ for consistency, which allows $(y_1,y_2,Y_3)$ to satisfy $y_1-2y_2+Y_3=3$. Since the column of the augmented matrix corresponding to $y_3$ will be upivoted, meaning that $y_3$ is a free variable in the solution set, the method will have an infinite number of solutions if this holds.
