Can't solve matrix multiplication with vector I try to solve this:
$$
\begin{bmatrix} 1-m & 0 & 0 \\ 0 & 2-m & 1 \\ 
 0 & 1 & 1-m\end{bmatrix}
\begin{bmatrix} h_{1}  \\ h_{2} \\ h_{3} \end{bmatrix} = 0
$$
That's what i tried:
=>
$$ (1-m) *h_1 = 0 $$
$$ (2-m) * h_{2} + h_{3} = 0 $$
$$ h_{2} + (1-m) * h_{3} = 0 $$
=>
$$ h_1 = 0 $$
$$ h_{3} = -((2-m) * h_{2}) $$
$$ h_{2} + (1-m) * h_{3} = 0 $$
=>
$$ h1 = 0 $$
$$ h_{3} = -2h_{2}+mh_{2} $$
$$ h_{2} + (1-m) * h_{3} = 0 $$
=> (Inserting h3 into equasion 3)
$$ h1 = 0 $$
$$ h_{3} = -2h_{2}+mh_{2} $$
$$ h_{2} + (1-m) * (-2h_{2}+mh_{2}) = 0 $$
=>
$$ h_1 = 0 $$
$$ h_{3} = -2h_{2}+mh_{2} $$
$$ h_{2} -2h_{2}+3mh_{2}-m^2h_{2} = 0 $$
=>
$$ h_1 = 0 $$
$$ h_{3} = -2h_{2}+mh_{2} $$
$$ -h_{2}+3mh_{2}-m^2h_{2} = 0 $$
But the result should be:
\begin{bmatrix} 0  \\ m-1 \\ 1 \end{bmatrix}
Can anyone show me what i`m doing wrong? Thanks in advance!

Clearly one solution is $0$ vector, that is $h_i=0$ for all $i$. If you want a solution different from  the $0$ the determinant of a given matrix must be $0$. So $$(1-m)((2-m)(1-m)-1)=0$$
so you get $m_1=1$ or $m^2-3m+1=0 \implies m_{2,3} = {3\pm \sqrt{5}\over 2}$
However in the first case we get $h_2=h_3=0$ and $h_1\in\mathbb{R}$ so the solution you mentioned clearly doesn't work. 
