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See the following vectors in $\mathbb{R}^3$:

$ \mathbf{u}_1= \begin{pmatrix} 1 \\ -1\\ 1 \\ 1 \end{pmatrix} $, $ \mathbf{u}_2 = \begin{pmatrix} 1 \\ 2 \\ 0 \\ 2 \end{pmatrix} $, $ \mathbf{u}_3 = \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} $

and

$ \mathbf{v}_1= \begin{pmatrix} 5 \\ -3\\ -4 \\ -1 \end{pmatrix} $, $ \mathbf{v}_2 = \begin{pmatrix} 6 \\ 4 \\ 1 \\ 8 \end{pmatrix} $, $ \mathbf{v}_3 = \begin{pmatrix} 7 \\ 2 \\ -1 \\ 6 \end{pmatrix} $

We are told that $\mathcal{B} = (\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3)$ and $\mathcal{C} = (\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3)$ both are bases for the same subspace $\mathcal{U}$.

Let $\mathcal{\lambda}$ be an unknown number og see the vector $\mathbf{x} = \mathbf{v}_1 + \lambda\mathbf{v}_2 - \mathbf{v}_3$.

Decide a number $a, b, c$ (expressed by $\lambda$) so that $a\mathbf{u}_1 + b\mathbf{u}_2 + c\mathbf{u}_3 = \mathbf{x}$

How do I go on about this?

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  1. Find some $\alpha_i$ such that $v_1=\alpha_1 u_1+\alpha_2u_2+\alpha_3u_3$
  2. Find some $\beta_i$ such that $v_2=\beta_1 u_1+\beta_2u_2+\beta_3u_3$
  3. Find some $\gamma_i$ such that $v_3=\gamma_1 u_1+\gamma_2u_2+\gamma_3u_3$
  4. Simplify the expression $$x=v_1+\lambda v_2-v_3=\alpha_1 u_1+\alpha_2u_2+\alpha_3u_3+\lambda(\beta_1 u_1+\beta_2u_2+\beta_3u_3)-(\gamma_1 u_1+\gamma_2u_2+\gamma_3u_3)$$
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  • $\begingroup$ Thanks 5xum! When I reduce it, I get $\begin{pmatrix} 8 \\ 3\\ -9 \\ 1 \end{pmatrix} + a\begin{pmatrix} 2 \\ -2\\ 1 \\ 1 \end{pmatrix} + b\begin{pmatrix} 3 \\ 6\\ 0 \\ 6 \end{pmatrix} + c\begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} $ What do I do from here? $\endgroup$ – jubibanna May 15 '19 at 13:56
  • $\begingroup$ I managed to solve it following your hint, thank you! $\endgroup$ – jubibanna May 15 '19 at 19:41

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