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Suppose the non zero $n \times 1$ column vector $x$ solves the system of equations $Ax=b$ where $A$ is $m \times n$ matrix whose columns are the vectors $a_1,a_2,\ldots,a_n$ and $b$ is a $m \times 1$ column vector. then the set of the vectors $\{a_1, a_2, \ldots, a_n, b\}$ is Linearly dependent . Please explain how? Shouldn't there be conditon the set of vectors in "$A$" matrix ?

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Note that $$Ax = b \iff \begin{bmatrix} a_1 & ... & a_n\end{bmatrix}\begin{bmatrix}x_1\\ \vdots \\ x_n\end{bmatrix} - b = 0 \iff \sum_{i=1}^n x_i a_i - b = \sum_{i=1}^n x_i a_i + (-1)\times b =0.$$

Since $-1\ne 0$, this proves that $(a_1,...,a_n,b)$ are linearly dependent.

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  • $\begingroup$ Please explain the last step, Since -1 not equal to zero, this gives desired relation $\endgroup$ – Anonymous May 10 '18 at 17:36
  • $\begingroup$ I edited my answer to be clearer. $\endgroup$ – paf May 10 '18 at 17:37
  • $\begingroup$ No, what I want to ask is how (-1) came ? $\endgroup$ – Anonymous May 10 '18 at 17:43
  • $\begingroup$ It's the coefficient of b, see my edit. $\endgroup$ – paf May 10 '18 at 17:44
  • $\begingroup$ Now it's clear ,Upvoted!! $\endgroup$ – Anonymous May 10 '18 at 17:48

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