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# For each of the following linearly dependent sets, find a redundant vector v in S and verify that $\operatorname{span}(S-\{\mathbf{v}\})=\operatorname{span}(S)$. (a) $S=\{[4,-2,6,1],[1,0,-1,2],[0,0,0,0],[6,-2,5,5]\}$, *(b) $S=\{[1,1,0,0],[1,1,1,0],[0,0,-6,0]\}$, (c) $S=\left\{\left[x_{1}, x_{2}, x_{3}, x_{4}\right] \in \mathbb{R}^{4} | x_{i}=\pm 1\right.$, for each $i \}$.

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Answer for (a): Given that $S$ be the linearly dependent set of vectors which contains 4 vectors

$S=\{[4,-2,6,1],[1,0,-1,2],[0,0,0,0],[6,-2,5,5]\}$

We prove that $\mathbf{v}=[0,0,0,0]$ be the redundant vector.
If we verify that $\text{span}(S-\{\mathbf{v}\})=\text{span}(S)$, then we can say that $\mathbf{v}=[0,0,0,0]$ is the redundant vector.
Now we apply the Simplified Span Method to both $(S-\{\mathbf{v}\})$ and $S$.
Applying the Span method on $(S-\{\mathbf{v}\})$ and $S$ if we have same row reduced echelon form (there may be an extra row of zeros) with $\mathbf{v}$ or with out $\mathbf{v}$ as one of the rows, then we can say that two spans are equal.

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