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$$ \vec a - \vec b + \vec c = \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 1 & 1\end{pmatrix} \begin{pmatrix} 1 \\-1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ $$ \vec a - \vec c + \vec d = \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 1 & 1\end{pmatrix} \begin{pmatrix} 1 \\0 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ Hopefully those are all clear, now I can multiply these both by $\alpha$ and $\beta$ respectively and add them like you describe, $$ \alpha( \vec a - \vec b + \vec c)+\beta( \vec a - \vec c + \vec d )= \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ $$ \alpha \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 1 & 1\end{pmatrix} \begin{pmatrix} 1 \\-1 \\ 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 1 & 1\end{pmatrix} \begin{pmatrix} 1 \\0 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ Scalar multiplication commutes with everything and we can factor out the matrix on the left of each, $$=\begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 1 & 1\end{pmatrix} \left[\alpha \begin{pmatrix} 1 \\-1 \\ 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 1 \\0 \\ -1 \\ 1 \end{pmatrix}\right] = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ Hopefully this helps to see it all out in full like this.
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Thu, 26 Jul 2018 08:29 GMT