Finding a General Solution to a Nonhomogeneous Matrix Equation I've come across a question in which I've been asked to find the general solution to the matrix equation: 
$\begin{bmatrix}1 & -2 & 1\\-2 & 4 &-2\\1 & -2 & 1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-16\\8\end{bmatrix}$
by first finding a solution to a similar homogeneous equation:
$\begin{bmatrix}1 & -2 & 1\\-2 & 4 &-2\\1 & -2 & 1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$
and then using that along with a given solution,
\begin{bmatrix}1\\-2\\3\end{bmatrix}
to find the general solution.
I've been looking around online and in textbooks for awhile and can't seem to find any information on this type of problem. Could someone provide a rundown of the methodology?
 A: General solution = (particular solution) + (general solution of the homogeneous equation).
A: Notice the column structure of the target matrix:
$$
\mathbf{A}=
%
\left[
\begin{array}{rrr}
 1 & -2 & 1 \\
 -2 & 4 & -2 \\
 1 & -2 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{rrr}
  c_{1} & -2 c_{1} & c_{1}
\end{array}
\right].
$$
We have one essential column. The rank plus nullity theorem reveals there will be two vectors to span the nullspace. 
$$
 \mathcal{N}\left( \mathbf{A} \right) = \text{span }
\left\{ \,
\left[
\begin{array}{r}
 2 \\
 1 \\
 0
\end{array}
\right], \,
\left[
\begin{array}{r}
 1 \\
 0 \\
 -1
\end{array}
\right] \,
\right\}
$$
For the data vector 
$$
b= 
\left[
\begin{array}{r}
 1 \\
 0 \\
 -1
\end{array}
\right] 
$$
The general solution for $\mathbf{A}x = b$ is
$$
  x = 
\left[
\begin{array}{c}
 8 \\
 0 \\
 0
\end{array}
\right] 
+
\alpha
\left[
\begin{array}{c}
 2 \\
 1 \\
 0
\end{array}
\right] 
+
\beta
\left[
\begin{array}{r}
  1 \\
  0 \\
 -1
\end{array}
\right], \qquad \alpha, \beta \in \mathbb{C}.
$$
