Solving a system of three linear equations using matrix operations Problem:
Solve the following system of linear equations.
with
\begin{align*} 
x + y + z &= 6 \\
x + 2y - z &= 2 \\
x + y + 2z &= 9 \\
\end{align*}
Answer:
\begin{align*} 
\begin{bmatrix}
1 & 1 & 1 & 6 \\
1 & 2 & -1 & 2 \\
1 & 1 & 2 & 9 \\
\end{bmatrix} \sim
\begin{bmatrix}
1 & 1 & 1 & 6 \\
0 & 1 & -2 & -4 \\
0 & 0 & 1 & 3
\end{bmatrix} \\
\end{align*}
Hence we have:
$z = 3$
Now we need to find $y$.
\begin{align*} 
y - 2z &= -4 \\
y - 2(3) &= -4 \\
y - 6 &= -4 \\
y &= 2 \\
\end{align*}
Now we need to find $x$.
\begin{align*} 
x + y + z &= 6
x + 2 + 3 &= 6 \\
x + 5 &= 6 \\
\end{align*}
Hence the solution is:
$$  x = 1, \,\, y = 2, \,\, z = 3 $$
Is my solution right? What bothers me is that I only needed two row operations. I am thinking this problem is particular easy.

Here is an independent confirmation of your solution. The system can be written in matrix form as $Ax=b$ with
$$
A=\begin{pmatrix} 1 & 1 & 1 \cr 1 & 2 & -1 \cr 1 & 1 & 2\end{pmatrix}.
$$
Since $\det(A)=1$, the matrix is invertible so that the unique solution is given by
$$x=A^{-1}b= \begin{pmatrix} 5 & -1 & -3 \cr -3 & 1 & 2 \cr -1 & 0 & 1\end{pmatrix} \begin{pmatrix} 6 \cr 2\cr 9\end{pmatrix}=\begin{pmatrix} 1 \cr 2\cr 3\end{pmatrix}.
$$
