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Check whether the following rows are linearly dependent/independent: \begin{aligned}(1,1,1,1,1,1,1,1,1,1),\\ (1,1,1,1,1,0,1,1,1,0),\\ (1,1,1,1,1,1,0,1,1,0),\\ (1,1,1,1,1,1,1,0,1,0),\\ (1,1,1,1,1,1,1,1,0,0),\\ (1,1,1,1,1,0,0,0,0,-3).\end{aligned}

If linearly dependent find the relation between them.

I tried to do using the combination $$c_1(1,1,1,1,1,1,1,1,1,1)+c_2(1,1,1,1,1,0,1,1,1,0)+c_3(1,1,1,1,1,1,0,1,1,0)\\+c_4(1,1,1,1,1,1,1,0,1,0)+c_5 (1,1,1,1,1,1,1,1,1,0)+c_6 (1,1,1,1,1,0,0,0,0,-3)\\=(0,0,0,0,0,0,0,0,0,0)$$

and I got $c_1=3c_6, \;c_2=c_3=c_4=c_5=0.$

But $c_1-3c_6\neq 0.$

So where am I wrong? Please help.

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  • $\begingroup$ I think for $c_6=1$ and $c_1=3$ you get what you need $\endgroup$ – dmtri Aug 29 '18 at 7:15
  • $\begingroup$ @PureMathematics Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Sep 17 '18 at 20:22
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Subtract every row (but the first) from the first.

$(1,1,1,1,1,1,1,1,1,1) \\(0,0,0,0,0,1,0,0,0,1) \\(0,0,0,0,0,0,1,0,0,1) \\(0,0,0,0,0,0,0,1,0,1) \\(0,0,0,0,0,0,0,0,0,1) \\(0,0,0,0,0,1,1,1,1,4)$

Then subtract row $2$ to $5$ from $6$.

$(1,1,1,1,1,1,1,1,1,1) \\(0,0,0,0,0,1,0,0,0,1) \\(0,0,0,0,0,0,1,0,0,1) \\(0,0,0,0,0,0,0,1,0,1) \\(0,0,0,0,0,0,0,0,0,1) \\(0,0,0,0,0,0,0,0,1,0)$

...

$(1,1,1,1,1,0,0,0,0,0) \\(0,0,0,0,0,1,0,0,0,0) \\(0,0,0,0,0,0,1,0,0,0) \\(0,0,0,0,0,0,0,1,0,0) \\(0,0,0,0,0,0,0,0,0,1) \\(0,0,0,0,0,0,0,0,1,0)$

This should be convincing enough. (As José did, you can dispense yourself with the first four columns.)

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Let put the vectors as rows in a matrix and check for the RREF. From the number of pivots we can deduce whether or not the vectors are linearly independent.

Refer also to Reduced row echelon form and linear independence

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I don't know where you got $c_1-3c_6=0$. The vectors are linearly independent. For instance if you forget the first $4$ coordinates of each vector and you obtain a $6\times6$ matrix from the rest, then that matrix is$$\begin{bmatrix}1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & -3\end{bmatrix},$$whose determinant is $1\neq0$.

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