Question

Alguém pode ajudar ,
Seja o espaço vetorial M(2,2) e os vetores 1 = 1 ,0
1 ,1
v2= −1, 2
0 ,1
3 = 0 −1
2 1 ,
escreva o vetor = 1 ,8
0, 5
como combinação linear dos vetores , v1 v2 e v3

Answer 1

Sendo uma combinação linear, devem existir escalares (números reais) $a_1$, $a_2$ e $a_3$ tal que:

$\begin{bmatrix}1 &8 \\ 0 &5 \end{bmatrix}=a_1\vec{v_1}+a_2\vec{v_2}+a_3\vec{v_3}$

$\begin{bmatrix}1 &8 \\ 0 &5 \end{bmatrix}=a_1\begin{bmatrix}1 &0 \\ 1 &1 \end{bmatrix}+a_2\begin{bmatrix}-1 &2 \\ 0 &1 \end{bmatrix}+a_3\begin{bmatrix}0 &-1 \\ 2 &1 \end{bmatrix}$

$\begin{bmatrix}1 &8 \\ 0 &5 \end{bmatrix}=\begin{bmatrix}a_1 &0 \\ a_1 &a_1 \end{bmatrix}+\begin{bmatrix}-a_2 &2a_2 \\ 0 &a_2 \end{bmatrix}+\begin{bmatrix}0 &-a_3 \\ 2a_3 &a_3 \end{bmatrix}$

$\begin{bmatrix}1 &8 \\ 0 &5 \end{bmatrix}=\begin{bmatrix}a_1-a_2 &2a_2-a_3 \\ a_1+2a_3 &a_1+a_2+a_3 \end{bmatrix}$

Daí tiramos o seguinte sistema:

$\left\{\begin{matrix}a_1-a_2=1\\2a_2-a_3=8\\a_1+2a_3=0\\a_1+a_2+a_3=5\end{matrix}\right.$

Reescrevendo o sistema na forma matricial:

$\begin{bmatrix}1 &-1 &0 \\ 0 &2 &-1 \\ 1 &0 &2\\ 1 &1 &1 \end{bmatrix}\left.\begin{matrix}1\\8\\0\\5\end{matrix}\right|$

Basta agora escalonarmos a matriz acima:

$\begin{bmatrix}1 &-1 &0 \\ 0 &2 &-1 \\ 1 &0 &2\\ 1 &1 &1 \end{bmatrix}\left.\begin{matrix}1\\8\\0\\5\end{matrix}\right|\xrightarrow[L_4-L_1\rightarrow L_4]{L_3-L_1\rightarrow L_3}\begin{bmatrix}1 &-1 &0 \\ 0 &2 &-1 \\ 0 &1 &2\\ 0 &2 &1 \end{bmatrix}\left.\begin{matrix}1\\8\\-1\\4\end{matrix}\right|\xrightarrow[]{1/2L_2\rightarrow L_2}\begin{bmatrix}1 &-1 &0 \\ 0 &1 &-1/2 \\ 0 &1 &2\\ 0 &2 &1 \end{bmatrix}\left.\begin{matrix}1\\4\\-1\\4\end{matrix}\right|$

$\begin{bmatrix}1 &-1 &0 \\ 0 &1 &-1/2 \\ 0 &1 &2\\ 0 &2 &1 \end{bmatrix}\left.\begin{matrix}1\\4\\-1\\4\end{matrix}\right|\xrightarrow[L_3-L_2\rightarrow L_3]{L_1+L_2\rightarrow L_1}\begin{bmatrix}1 &0 &-1/2 \\ 0 &1 &-1/2 \\ 0 &0 &5/2\\ 0 &2 &1 \end{bmatrix}\left.\begin{matrix}5\\4\\-5\\4\end{matrix}\right|$

$\begin{bmatrix}1 &0 &-1/2 \\ 0 &1 &-1/2 \\ 0 &0 &5/2\\ 0 &2 &1 \end{bmatrix}\left.\begin{matrix}5\\4\\-5\\4\end{matrix}\right|\xrightarrow[2/5L_3\rightarrow L_3]{L_4-2L_2\rightarrow L_4}\begin{bmatrix}1 &0 &-1/2 \\ 0 &1 &-1/2 \\ 0 &0 &1\\ 0 &0 &2 \end{bmatrix}\left.\begin{matrix}5\\4\\-2\\-4\end{matrix}\right|$

$\begin{bmatrix}1 &0 &-1/2 \\ 0 &1 &-1/2 \\ 0 &0 &1\\ 0 &0 &2 \end{bmatrix}\left.\begin{matrix}5\\4\\-2\\-4\end{matrix}\right|\xrightarrow[L_2+1/2L_3\rightarrow L_2]{L_4-2L_1\rightarrow L_4}\begin{bmatrix}1 &0 &-1/2 \\ 0 &1 &0 \\ 0 &0 &1\\ 0 &0 &0 \end{bmatrix}\left.\begin{matrix}5\\3\\-2\\0\end{matrix}\right|\xrightarrow[]{L_1+1/2L_3\rightarrow L_1}\begin{bmatrix}1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1\\ 0 &0 &0 \end{bmatrix}\left.\begin{matrix}4\\3\\-2\\0\end{matrix}\right|$Com isso concluímos que $a_1=4$, $a_2=3$ e $a_3=-2$, logo:

$\begin{bmatrix}1 &8 \\ 0 &5 \end{bmatrix}=4\begin{bmatrix}1 &0 \\ 1 &1 \end{bmatrix}+3\begin{bmatrix}-1 &2 \\ 0 &1 \end{bmatrix}-2\begin{bmatrix}0 &-1 \\ 2 &1 \end{bmatrix}$