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I was reading Vector Linear Independence where I came across a matrix operation I have never seen before. What is this operation called and is it generally valid? I mean there is a $2\times3$ matrix that was reduced to $2\times2$...!

$$\begin{bmatrix}1&0&\frac{16}{5}\\ 0 & 1 & \frac 25\end{bmatrix}\left\{\begin{array}{c}a_1\\a_2\\a_3\end{array}\right\}=\left\{\begin{array}{c}0\\0\end{array}\right\}$$ We can now rearrange this equation to obtain

$$\begin{bmatrix}1&0\\0&1\end{bmatrix}\left\{\begin{array}{c}a_1\\a_2\end{array}\right\}=\left\{\begin{array}{c}a_1\\a_2\end{array}\right\}=-a_3\left\{\begin{array}{c}\frac{16}{5}\\\frac{2}{5}\end{array}\right\}$$

Thank you in advance.

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    $\begingroup$ I'd say that's just rearrangement, nothing more, nothing less. $\endgroup$ – Cameron Williams Mar 8 '17 at 14:40
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    $\begingroup$ It's not reduced, it's just rewritten. $\endgroup$ – mathreadler Mar 8 '17 at 14:42
  • $\begingroup$ @CameronWilliams, correct, I now get it. Thx. $\endgroup$ – NoChance Mar 8 '17 at 14:54
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There is nothing special. We have $$ \begin{bmatrix} 1 & 0 & 16/5 \\ 0 & 1 & 2 /5 \end{bmatrix} \cdot \begin{bmatrix} a_1\\a_2\\a_3 \end{bmatrix} = \color{blue}{\begin{bmatrix} 1 \\ 0 \end{bmatrix}\cdot a_1 + \begin{bmatrix} 0\\ 1 \end{bmatrix} \cdot a_2} + \begin{bmatrix} 16/5\\ 2/5 \end{bmatrix}\cdot a_3 $$ and the blue part can be rewritten as $$ \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix} \cdot \begin{bmatrix} a_1\\a_2 \end{bmatrix} $$

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  • $\begingroup$ Coloring helped! Thank you for your effort. $\endgroup$ – NoChance Mar 8 '17 at 14:45

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