Problem: Let
\[
A =
\begin{bmatrix}
3 & -1 & 2 \\
2 & 1 & 1 \\
1 & -3 & 0 \\
\end{bmatrix}.
\]
For which triples $(y_1, y_2, y_3)$ does the system $AX = Y$ have a solution?
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Solution: Consider the system $AX = Y$, where $Y = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}$. Let us do the row-reduction of the augmented matrix.
\begin{align*}
\begin{bmatrix}
3 & -1 & 2 & y_1\\
2 & 1 & 1 & y_2\\
1 & -3 & 0 & y_3\\
\end{bmatrix}
& \xrightarrow{
\begin{matrix}
R_1 \leftrightarrow \frac{1}{3}R_1
\end{matrix}
}
\begin{bmatrix}
1 & -\frac{1}{3} & \frac{2}{3} & \frac{y_1}{3} \\[1ex]
2 & 1 & 1 & y_2 \\
1 & -3 & 0 & y_3 \\
\end{bmatrix} \\[2ex]
& \xrightarrow{
\begin{matrix}
R_2 \rightarrow R_2 - 2R_1 \\
R_3 \rightarrow R_3 - R_1
\end{matrix}
}
\begin{bmatrix}
1 & -\frac{1}{3} & \frac{2}{3} & \frac{y_1}{3} \\[1ex]
0 & \frac{5}{3} & -\frac{1}{3} & y_2 - \frac{2y_1}{3} \\[1ex]
0 & -\frac{8}{3} & -\frac{2}{3} & y_3 - \frac{y_1}{3} \\
\end{bmatrix}\\[2ex]
& \xrightarrow{
\begin{matrix}
R_2 \leftrightarrow \frac{3}{5}R_2
\end{matrix}
}
\begin{bmatrix}
1 & -\frac{1}{3} & \frac{2}{3} & \frac{y_1}{3} \\[1ex]
0 & 1 & -\frac{1}{5} & \frac{3y_2 - 2y_1}{5} \\[1ex]
0 & -\frac{8}{3} & -\frac{2}{3} & y_3 - \frac{y_1}{3} \\
\end{bmatrix}\\[2ex]
& \xrightarrow{
\begin{matrix}
R_1 \rightarrow R_1 + \frac{1}{3}R_2 \\[0.5ex]
R_3 \rightarrow R_3 + \frac{8}{3}R_2
\end{matrix}
}
\begin{bmatrix}
1 & 0 & \frac{3}{5} & \frac{1}{5}y_1 + \frac{1}{5}y_2 \\[1ex]
0 & 1 & -\frac{1}{5} & \frac{3y_2 - 2y_1}{5} \\[1ex]
0 & 0 & -\frac{6}{5} & -\frac{7}{5}y_1 + \frac{8}{5}y_2 + y_3 \\
\end{bmatrix}\\[2ex]
& \xrightarrow{
\begin{matrix}
R_3 \leftrightarrow -\frac{5}{6}R_3
\end{matrix}
}
\begin{bmatrix}
1 & 0 & \frac{3}{5} & \frac{1}{5}y_1 + \frac{1}{5}y_2 \\[1ex]
0 & 1 & -\frac{1}{5} & \frac{3y_2 - 2y_1}{5} \\[1ex]
0 & 0 & 1 & \frac{7}{6}y_1 - \frac{4}{3}y_2 - \frac{5}{6}y_3 \\
\end{bmatrix}\\[2ex]
& \xrightarrow{
\begin{matrix}
R_1 \rightarrow R_1 - \frac{3}{5}R_3 \\[0.5ex]
R_2 \rightarrow R_2 + \frac{1}{5}R_3
\end{matrix}
}
\begin{bmatrix}
1 & 0 & 0 & -\frac{1}{2}y_1 + y_2 + \frac{1}{2}y_3 \\[1ex]
0 & 1 & 0 & -\frac{1}{6}y_1 + \frac{1}{3}y_2 - \frac{1}{6}y_3 \\[1ex]
0 & 0 & 1 & \frac{7}{6}y_1 - \frac{4}{3}y_2 - \frac{5}{6}y_3 \\
\end{bmatrix}\\[2ex]
\end{align*}
Since every row contains a nonzero entry in the first three columns, the system of equations $AX = Y$ is consistent for any values of $y_1, y_2, y_3$. Therefore, the system $AX = Y$ has a unique solution for any values of $y_1, y_2, y_3$.