Resposta :
Olá, Eric.
Devemos encontrar [tex]\lambda_1,\lambda_2[/tex] tais que:
[tex]u=\lambda_1v_1+\lambda_2v_2 \Rightarrow \\\\ 2i-3j=\lambda_1 (-2i -4j)+\lambda_2( 3i-j) \Rightarrow \\\\ 2i-3j=-2\lambda_1i -4\lambda_1j+3\lambda_2i-\lambda_2j \Rightarrow \\\\ 2i-3j=(-2\lambda_1+3\lambda_2)i +(-4\lambda_1-\lambda_2)j \Rightarrow \\\\ \begin{cases} -2\lambda_1+3\lambda_2=2\\ -4\lambda_1-\lambda_2=-3\ \ (\times 3)\end{cases} \Rightarrow \begin{cases} -2\lambda_1+3\lambda_2=2\\ -12\lambda_1-3\lambda_2=-9\end{cases}\ \bigoplus\Rightarrow[/tex]
[tex]-14\lambda_1=-7 \Rightarrow \boxed{\lambda_1=\frac12} \Rightarrow [/tex]
[tex]-1+3\lambda_2=2 \Rightarrow \boxed{\lambda_2=1}[/tex]
[tex]\therefore u=\lambda_1v_1+\lambda_2v_2 \Rightarrow \boxed{u=\frac12 v_1+v_2}[/tex]