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I need help expressing \begin{bmatrix}6\\4\\3\end{bmatrix}

as a linear combination of the vectors $$ \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} \text{ , } \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} \text{ and } \begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix} $$

My approach was to create the augmented matrix $$ \left[\begin{array}{rrr|r} 1&3&3&6\\2&2&1&4\\-1&-1&2&3\end{array}\right] $$

and row reduce it to find x1, x2, and x3. Then I will write x1[matrix1] + x2[matrix2] + x3[matrix3]. Is this the correct approach?

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    $\begingroup$ Looks good. Just be aware of how to tell if there is no solution! $\endgroup$ – The Count Jan 30 '17 at 2:32
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Let v = $\begin{bmatrix} 6 \\ 4 \\ 3 \end{bmatrix}$, x = $\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$, y = $\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}$, and $z = \begin{bmatrix} 3 \\ 2 \\ -1 \end{bmatrix}$. So, we want to find real numbers a, b, and c such that v = ax + by + cz. This leaves you with a set of three equations in three unknowns. $$ 6 = a + 3b + 3c$$ $$4 = 2a + b + 2c$$ $$3 = -a + 2b - c $$ If you solve this system using any method you like, you'll see that {a, b, c} = {$\frac{3}{2}$, 2, $\frac{-1}{2}$}

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    $\begingroup$ Yes, thank you for verification. $\endgroup$ – user400359 Jan 30 '17 at 4:15
  • $\begingroup$ @stackofhay42 Could you please upvote me? $\endgroup$ – user7348 Jan 30 '17 at 15:49

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