The coefficient matrix of this system of equations is
\begin{equation*}
A=
\begin{bmatrix}
1 & 2 & -1 \\
2 & 5 & -1 \\
-1 & -4 & 0
\end{bmatrix}
\end{equation*}
and the vector of constants is $\vect{b}=\colvector{-3\\-4\\2}$. So by Theorem SLEMM we can convert the system to the form $A\vect{x}=\vect{b}$. Row-reducing this matrix yields the identity matrix so by Theorem NMRRI we know $A$ is nonsingular. This allows us to apply Theorem SNCM to find the unique solution as
\begin{equation*}
\vect{x}
=
\inverse{A}\vect{b}
=
\begin{bmatrix}
-4 & 4 & 3 \\
1 & -1 & -1 \\
-3 & 2 & 1
\end{bmatrix}
\colvector{-3\\-4\\2}
=
\colvector{2\\-1\\3}
\end{equation*}
Remember, you can check this solution easily by evaluating the matrix-vector product $A\vect{x}$ (Definition MVP).