Can't solve matrix multiplication with vector

Question

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!

Answer 1

HINT

For $m\neq1,2$ from here

• $ (1-m) h_1 = 0 \implies h_1=0$

• $ (2-m) h_{2} + h_{3} = 0 $

• $ h_{2} + (1-m) h_{3} = 0 \implies (2-m) h_{2} + (2-m) (1-m) h_{3} = 0$

then subtract the second and the third to find $h_3$, then find $h_2$ from the second.

Finally consider a part the cases with $m=1$ and $m=2$.

Answer 2

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.