Can $(0,0,0)$ be considered a particular solution in linear algebra? For the following matrix $A$:
$$
\left[    
\begin{matrix}
    1 &amp; 5 &amp; 2 \\
    \end{matrix}
\right]$$
If we solve $Ax = 0$, we get the general solution as:
$$
\left[
\begin{matrix}
-5t_1 - 2t_2\\
t_1\\
t_2
\end{matrix}
\right]
$$
Is $(0,0,0)$ a particular solution or does a particular solution not exist?
In general solutions for a linear system are given by
$$Ax=b \quad x=x_P+x_H$$
such that
$$Ax_P=b \quad Ax_H=0 \implies A(x_P+x_H)= Ax_P+Ax_H=b+0=b$$
In this case $b=0$ therefore the zero vector is a particular solution and some not trivial particular solution are for example
$$\left[
\begin{matrix}
0\\
\frac15\\
-\frac12
\end{matrix}
\right],\quad \left[
\begin{matrix}
1\\
0\\
-\frac12
\end{matrix}
\right], \quad \left[
\begin{matrix}
1\\
-\frac15\\
0
\end{matrix}
\right]$$
In general solutions for a linear system are given by
$$Ax=b \quad x=x_P+x_H$$
such that
$$Ax_P=b \quad Ax_H=0 \implies A(x_P+x_H)= Ax_P+Ax_H=b+0=b$$
In this case $b=0$ therefore the zero vector is a particular solution and some not trivial particular solution are for example
$$\left[
\begin{matrix}
0\\
\frac15\\
-\frac12
\end{matrix}
\right],\quad \left[
\begin{matrix}
1\\
0\\
-\frac12
\end{matrix}
\right], \quad \left[
\begin{matrix}
1\\
-\frac15\\
0
\end{matrix}
\right]$$