Show that the equation Ax=x can be rewritten as (A-I)x = 0 and use this result to solve Ax=x for x. Given matrix A = \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix}
and x = \begin{bmatrix}x_1  \\ x_2 \\ x_3 \end{bmatrix}
Answer: Given $Ax = x$. Subtract from x from both sides to get $Ax - x = 0$. We know $x =  I \cdot  x$ , where $I$ is the identity matrix.
Simplify $Ax - x = 0$ to get $x(A - I) = 0$.
$$\begin{bmatrix}x_1  \\ x_2 \\ x_3 \end{bmatrix} \cdot  \left(\begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \right) = \begin{bmatrix}0  \\ 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix}x_1  \\ x_2 \\ x_3 \end{bmatrix} \cdot \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}0  \\ 0 \\ 0 \end{bmatrix}$$
Do I need to write the system of linear equations?
 A: First it is important, that $\textbf A \textbf x - \textbf x = \textbf 0 \Rightarrow (\textbf A - \textbf I)\textbf x=0$. The vector $\textbf x$ has to be on the right side of $(\textbf A-\textbf I)$. Otherwise it wouldn´t work. And you have to subtract $\textbf I$ from $\textbf A$.
$$ \left( \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix} -\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \right) \cdot  \begin{bmatrix}x_1  \\ x_2 \\ x_3 \end{bmatrix}  = \begin{bmatrix}0  \\ 0 \\ 0 \end{bmatrix}$$
After calculating the brackets you can write a linear system of equations.
A: One has
\begin{equation}
A-I
= 
\left(\begin{matrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{matrix}\right)
-
\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)
=
\left(\begin{matrix}1 & 1 & 2 \\ 2 & 1 & -2 \\ 3 & 1 & 0\end{matrix}\right)
\end{equation}
With the rule of Sarrus
(see http://en.wikipedia.org/wiki/Rule_of_Sarrus), one gets
\begin{equation}
det(A-I) = 0 + (-6) + (+4) - (-2) - 0 - (+6) = -6
\end{equation}
Thus one has $x=0$.$\Box$
