If and when a linear system has exactly three solutions 
Does the following system have exactly three solutions?
  $$\left\{ \begin{array}{l}
2x - y +3z = 1 \\
x + 4y - 2z = -7 \\
3x + y -z = 4 \\
\end{array} \right.$$

I marked this answer as True. 
I proceeded to row reduce it and obtained the resultant matrix as - 
$$\left[ \begin{array}{ccc|c}
1 & -1/2 & 3/2 & 4 \\
0 & 1 & 9 & 13/9 \\
0 & 0 & 1 & 10/28 \\
\end{array} \right]$$
I'm pretty sure I made some mistake while row-reducing it. I answered this question in an exam setting. Then I gave the explanation as follows - 
Form row reduction, we know that the system is consistent and the rank of the matrix is 3 which is equal to the number of variables. So, the system of equations has 3 distinct solutions. 
Can someone point out where I went wrong? 
 A: Using row reduction you can show that $x=7/4$, $y=-25/8$ and $z=-15/8$. You can use even Cramer rule to compute the unique solution using that the matrix $A$ given by 
$$A=\left(\begin{array}{ccc} 2&-1&3\\  1&4&-2\\ 3&1&-1\end{array}\right)$$
and $\det(A)=-32\neq 0$ and this implies that the system has unique solution for any vector of independent terms. 
A: HINT: The system of linear equations can have only 0, 1 or infinitely many solutions. Hence no calculations are needed.
A: Start with the augmented matrix:
\begin{eqnarray}
\left(\begin{array}{cccc} 2&-1&3 &: 1\\  1&4&2&:-7\\ 3&1&-1&:4\end{array}\right)
\end{eqnarray}
First: Interchange rows 1 and 2 to get:
$$\left(\begin{array}{cccc}  1&4&-2&:-7\\  2&-1&3 &: 1\\ 3&1&-1&:4\end{array}\right)$$
Second: Use the first element of the first row as a pivot to eliminate the 2 and 3 of the first column to get:
$$ \left(\begin{array}{cccc}  1&4&-2&:-7\\  0&-9&7 &: 15\\ 0&-11&5&:25\end{array}\right) $$
Then you can replace Row2 by Row2-Row3 to get:
$$\left(\begin{array}{cccc}  1&4&-2&:-7\\  0&2&2&: -10\\ 0&-11&5&:25\end{array}\right)$$
then you can divide by 2 the Row2 to get:
$$\left(\begin{array}{cccc}  1&4&-2&:-7\\  0&1&1&: -5\\ 0&-11&5&:25\end{array}\right)$$
Then you can eliminate the -11 in the third row:
$$\left(\begin{array}{cccc}  1&4&-2&:-7\\  0&1&1&: -5\\ 0&0&16&:-30\end{array}\right)$$
So $z=\frac{-30}{16}=\frac{-15}{8}$. Because you have the equivalent augmentedmatrix
$$\left(\begin{array}{cccc}  1&4&-2&:-7\\  0&1&1&: -5\\ 0&0&1&:-\frac{15}{8}\end{array}\right)$$
To get $y$ you simply replace Row2 with Row2-Row3 and you will get:
$$\left(\begin{array}{cccc}  1&4&-2&:-7\\  0&1&0&: -5+\frac{15}{8}\\ 0&0&1&:-\frac{15}{8}\end{array}\right)$$
and $y=-5+\frac{15}{8}=\frac{-40+15}{8}=\frac{-25}{8}$. I leave $x$ for you to compute.
