Question

Verifique se o vetor v=(7,8,9) é combinacao linear dos vetores v1 (2,1,4) v2 (1, -1,3) v3 ( 3,2,5).

Answer 1

O vetor v=(7,8,9) é sim a combinação linear dos vetores v1 (2,1,4) v2 (1, -1,3) v3 ( 3,2,5).

• Dado que:

$\large\displaystyle\begin{aligned} \overset{\rightarrow}{V} = C_1 \overset{\rightarrow}{V_1} + C_2\overset{\rightarrow}{V_2} + C_3\overset{\rightarrow}{V_3} \Leftrightarrow\end{aligned}$

$\large\displaystyle\begin{aligned} \left[\begin{array}{ccc}7\\8\\9\end{array}\right] = C_1\left[\begin{array}{ccc}2\\1\\4\end{array}\right] + C_2\left[\begin{array}{ccc}1\\-1\\3\end{array}\right] + C_3\left[\begin{array}{ccc}3\\2\\5\end{array}\right] \Leftrightarrow\end{aligned}$

$\large\displaystyle\begin{aligned} \left[\begin{array}{ccc}7\\8\\9\end{array}\right] = \left[\begin{array}{ccc}2C_1\\1C_1\\4C_1\end{array}\right] + \left[\begin{array}{ccc}1C_2\\-1C_2\\3C_2\end{array}\right] + \left[\begin{array}{ccc}3C_3\\2C_3\\5C_3\end{array}\right] \Leftrightarrow\end{aligned}$

• Com isso, devemos criar um sistema linear de ordem 3, para assim encontrar o C₁ , C₂ e o C₃. Logo:

$\large\displaystyle\begin{aligned} \begin{cases} 2C_1 + C_2 + 3C_3 = 7\\ C_1 - 1C_2 + 2C_3=8\\ 4C_1 + 3C_2 + 5C_3 = 9\end{cases}\end{aligned}$

Resolvendo esse SL, iremos encontrar que C₁ = 0 ; C₂ = -2 ; C₃ = 3. Logo, iremos substituir e ver se a igualdade está realmente certa né.

$\large\displaystyle\begin{aligned} \left[\begin{array}{ccc}7\\8\\9\end{array}\right] = C_1\left[\begin{array}{ccc}2\\1\\4\end{array}\right] + C_2\left[\begin{array}{ccc}1\\-1\\3\end{array}\right] + C_3\left[\begin{array}{ccc}3\\2\\5\end{array}\right] \Leftrightarrow\end{aligned}$

$\large\displaystyle\begin{aligned} \left[\begin{array}{ccc}7\\8\\9\end{array}\right] = 0\cdot \left[\begin{array}{ccc}2\\1\\4\end{array}\right] + -2\cdot \left[\begin{array}{ccc}1\\-1\\3\end{array}\right] + 3\cdot \left[\begin{array}{ccc}3\\2\\5\end{array}\right] \Leftrightarrow\end{aligned}$

$\large\displaystyle\begin{aligned} \left[\begin{array}{ccc}7\\8\\9\end{array}\right] = \left[\begin{array}{ccc}0\\0\\0\end{array}\right] + \left[\begin{array}{ccc}-2\\2\\-6\end{array}\right] + \left[\begin{array}{ccc}9\\6\\15\end{array}\right] \Leftrightarrow\end{aligned}$

$\large\displaystyle\begin{aligned} \left[\begin{array}{ccc}7\\8\\9\end{array}\right] = \left[\begin{array}{ccc}0+(-2)+9\\0+2+6\\0+(-6)+15\end{array}\right] \Leftrightarrow\end{aligned}$

$\large\displaystyle\begin{aligned} \Leftrightarrow \left[\begin{array}{ccc}7\\8\\9\end{array}\right] = \left[\begin{array}{ccc}7\\8\\9\end{array}\right] \end{aligned}$

Perceba que a igualdade é verdadeira. Portanto o vetor v=(7,8,9) é a combinação linear dos vetores v1 (2,1,4) v2 (1, -1,3) v3 ( 3,2,5).

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