Non-trivial solution to a homogeneous system of linear equations. I have equations:
\begin{cases}
    2x + y - z = 0\\
    x - 2y - 3z =0\\
    -3x - y + 2z =0
\end{cases}
After I put this in matrix row-reduced echelon form I get solutions of $x=0, y=0, z=0$.  But my book says it has a non-trivial solution.  Could someone explain how that could be?
 A: Let $$A=
\begin{pmatrix}2 & 1 & -1\\1 & -2 & -3\\ -3 & -1 & 2  \end{pmatrix},$$
and $v_i$, $i=1,2,3$ be the column vectors of $A$. Observe that $v_1-v_2+v_3=0$. This implies that $\{v_1,v_2,v_3\}$ is linearly independent, so $\ker A$ is nontrivial. In particular, $$\begin{pmatrix}1\\-1\\1\\\end{pmatrix}\in\ker A.$$
A: We can write the LHS of your system as
$$
\left[
\begin{array}{rrr}
2 & 1 & - 1 \\
1 & -2 & - 3 \\
-3 & -1 &2
\end{array}
\right]
$$
and apply Gauß elimination:
$$
\to
\left[
\begin{array}{rrr}
1 & -2 & - 3 \\
2 & 1 & - 1 \\
-3 & -1 &2
\end{array}
\right]
\to
\left[
\begin{array}{rrr}
1 & -2 & - 3 \\
0 & 5 & 5 \\
0 & -7 & 7
\end{array}
\right]
\to
\left[
\begin{array}{rrr}
1 & -2 & - 3 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{array}
\right]
\to
\left[
\begin{array}{rrr}
1 & 0 & -1 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{array}
\right]
$$
So the solution can be written e.g. as
$$
L 
= \{ (z, -z, z) \mid z \in \mathbb{R} \}
= \{ z \, (1, -1, 1) \mid z \in \mathbb{R} \}
$$
A: The rank of the reduced row echelon matrix = the rank of the augmented matrix and this rank is less than the number of variables (unknown terms).
This system of linear equations has infinite number of solutions.
