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I am learning about Linear dependent vector from here

But I am unable to grasp the following equation:

If no such scalars exist, then the vectors are to be linearly independent.

$$c_1\begin{bmatrix}x_{11}\\x_{21}\\\vdots\\x_{n1}\\ \end{bmatrix}+c_2\begin{bmatrix}x_{12}\\x_{22}\\\vdots\\x_{n2}\\ \end{bmatrix}+\cdots+c_n\begin{bmatrix}x_{1n}\\x_{2n}\\\vdots\\x_{nn}\\ \end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\0\\ \end{bmatrix}\\ \begin{bmatrix}x_{11}&x_{12}&\cdots&x_{1n}\\x_{21}&x_{22}&\cdots&x_{2n}\\ \vdots&\vdots&\ddots&\vdots\\x_{n1}&x_{n2}&\cdots&x_{nn}&\\ \end{bmatrix}\begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\0\end{bmatrix}$$ In order for this matrix equation to have a nontrivial solution, the determinant must be $0$

How the first equation is reduced to the second one?

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  • $\begingroup$ What exactly are you asking? $\endgroup$ – user418131 Sep 18 '18 at 9:05
  • $\begingroup$ How the first equation is reduced to the second one? $\endgroup$ – Cody Sep 18 '18 at 9:06
  • $\begingroup$ Write a formula for the $i$-th entry of the vector above and below and you will see they are the same. $\endgroup$ – Michal Adamaszek Sep 18 '18 at 9:08
  • $\begingroup$ By the use of matrix multiplication. $\endgroup$ – user418131 Sep 18 '18 at 9:08
  • $\begingroup$ The first equation is actually equivalent to $n$ equations. Do you know how to interchange linear equations with an analogous matrix equation? $\endgroup$ – user418131 Sep 18 '18 at 9:10
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I hope this small example can help you to understand.

$$\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}\begin{bmatrix} c_{1} \\ c_{2}\end{bmatrix}=\begin{bmatrix} c_{1}x_{11}+c_2x_{12} \\ c_{1}x_{21}+c_2x_{22} \end{bmatrix}=c_1\begin{bmatrix} x_{11} \\ x_{21}\end{bmatrix}+ c_2\begin{bmatrix} x_{21} \\ x_{22}\end{bmatrix}$$

Notice that $c_1$ is only multiplied to entries in the first column and $c_2$ is only multiplied to the entries in the second column.

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This results from the definition of scalar multiplication and addition of matrices: \begin{align} c_1\begin{bmatrix} x_{11}\\x_{21}\\\vdots\\x_{n1} \end{bmatrix}+c_2\begin{bmatrix} x_{12}\\x_{22}\\\vdots\\x_{n2} \end{bmatrix}+\dots +c_n\begin{bmatrix} x_{1n}\\x_{2n}\\\vdots\\x_{nn} \end{bmatrix} &= \begin{bmatrix} c_1x_{11}\\c_1x_{21}\\\vdots\\c_1x_{n1} \end{bmatrix}+\begin{bmatrix} c_2x_{12}\\c_2x_{22}\\\vdots\\c_2x_{n2} \end{bmatrix}+\dots +\begin{bmatrix} c_nx_{1n}\\c_nx_{2n}\\\vdots\\c_nx_{nn} \end{bmatrix}\\[1ex] &= \begin{bmatrix} c_1x_{11}+c_2x_{12}+\dots+c_nx_{1n}\\c_1x_{21}+c_2x_{22}+\dots+c_nx_{2n}\\\dots\dots\dots\dots\dots\dots\dots\dots\\c_1x_{n1}+c_2x_{n2}+\dots+c_nx_{nn} \end{bmatrix} \end{align}

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  • $\begingroup$ So we are essentially writing a set of linear equations in matrix form, right? $\endgroup$ – Cody Sep 18 '18 at 9:54
  • $\begingroup$ It's exactly that. $\endgroup$ – Bernard Sep 18 '18 at 10:00
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Recall that the product $A\vec c$ can be interpreted as the linear combination of the colums $\vec x_i$ of $A$ by the coordinates $c_i$ of $\vec c$

$$A\vec c =\sum c_i\vec x_i$$

Refer also to the related

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