Linear Algebra: solution of homogeneous system of equation Could someone please explain to me how they found $x_1$? By my calculations I got $x_2$ & $x_3=0$ and $0=0$.

 A: The system says that $x_2$ must equal $0$, and $x_3$ must equal $0$; however, there are no constraints on $x_1$ (the third equation you got, which is "$0=0$", is always satisfied, so it imposes no conditions whatsoever on the solutions).
That means that for $\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)$ to be a solution, you need $x_2=0$, $x_3=0$, but $x_1$ can be anything; that is, the solutions are all vectors of the form
$$\left(\begin{array}{c}t\\0\\0\end{array}\right),$$
with $t$ arbitrary.
A: Well, if you carry out the matrix multiplication, you get:
$$
\begin{pmatrix}
0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}
=
\begin{pmatrix} x_2 \\ x_3 \\ 0 \end{pmatrix}
$$
for this to be equal to the zero vector, you indeed need $x_2 = 0$ and $x_3 = 0$. So far so good.
Now, notice that $x_1$ is not in any element of the vector on the right hand side, so there are no constraints on $x_1$. Therefore, $x_1$ can be any number, here called $t$. 
You can check your solution; do the multplication
$$
\begin{pmatrix}
0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix} t \\ 0 \\ 0 \end{pmatrix}
$$
and see what you get.
