a)

$\overrightarrow{u}=\overrightarrow{v}$  

$\left[m^2-1,-m\right]=\left[-1,m\right]$      

$\{\begin{array}{l}{m}^2-1=-1\\ -m=m\end{array}$  

$\{\begin{array}{l}{m}^2=0\\ -2m=0\end{array}$  

$m=0$  

---

b)

$\overrightarrow{u}=-\overrightarrow{v}$  

$\left[m^2-1,-m\right]=-\left[-1,m\right]$  

$\left[m^2-1,-m\right]=\left[1,-m\right]$  

$\{\begin{array}{l}{m}^2-1=1\\ -m=-m\end{array}$  

$\{\begin{array}{l}{m}^2-2=0\\ 0=0\end{array}$  

$m=\sqrt{2}\vee m=-\sqrt{2}$  

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c)

$\overrightarrow{u}=k\cdot \overrightarrow{v}$  

$\left[m^2-1,-m\right]=k\cdot \left[-1,m\right]$  

$\left[m^2-1,-m\right]=\left[-k,km\right]$  

$\{\begin{array}{l}{m}^2-1=-k\\ -m=km\mid +m\end{array}$  

$\{\begin{array}{l}{m}^2-1=-k\\ 0=km+m\end{array}$  

$\{\begin{array}{l}{m}^2-1=-k\\ 0=m\left(k+1\right)\end{array}$  

 

Przypadek I.

$m=0$  

 

Przypadek II. 

$k+1=0$  

$k=-1$  

Wtedy pierwsze równanie jest postaci:

$m^2-1=1$  

$m^2-2=0$  

$m=\sqrt{2}\vee m=-\sqrt{2}$