Answer
Does not have any solutions.
Work Step by Step
Consider the given system of equations:
$\begin{align}
& 2x-3y+z=1 \\
& x-2y+3z=2 \\
& 3x-4y-z=1 \\
\end{align}$
Therefore, in matrix form the system of equations can be written as below:
$ AX=b $
Where
$ A=\left[ \begin{array}{*{35}{r}}
2 & -3 & 1 \\
1 & -2 & 3 \\
3 & -4 & -1 \\
\end{array} \right];b=\left[ \begin{array}{*{35}{r}}
1 \\
2 \\
1 \\
\end{array} \right];X=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]$
Consider the augmented matrix
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
2 & -3 & 1 & 1 \\
1 & -2 & 3 & 2 \\
3 & -4 & -1 & 1 \\
\end{array} \right]$
Apply elementary row operation on $ A $ to convert it to its equivalent upper triangular matrix form.
Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-\frac{1}{2}{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-\frac{3}{2}{{R}_{1}}$ to get, $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
2 & -3 & 1 & 1 \\
0 & -\frac{1}{2} & \frac{5}{2} & \frac{3}{2} \\
0 & \frac{1}{2} & -\frac{5}{2} & -\frac{1}{2} \\
\end{array} \right]$
Step 2: Apply the operation ${{{R}'}_{3}}={{R}_{3}}+{{R}_{2}}$ to get, $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
2 & -3 & 1 & 1 \\
0 & -\frac{1}{2} & \frac{5}{2} & \frac{3}{2} \\
0 & 0 & 0 & 1 \\
\end{array} \right]$
Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is 3 and the rank of the coefficient matrix $ A $ is 2 -- that is $\text{rank}\left[ A|b \right]\ne \text{rank}\left[ A \right]$
