How to determine linear dependence

In $\mathbb R^3$, consider the following statements about the subset $$E = \{(1,0,0),(0,1,0),(0,0,1),(1,1,1),(1,1,0)\}$$

Which of the following is/are correct:

1. $E$ is linearly dependent.
2. Any three vectors in $E$ are linearly independant.
3. Any four vectors in $E$ are linearly dependent.

My attempt : (1,1,1) = (1,0,0)+(0,1,0)+(0,0,1), therefore, 1. is true.

Also, since $\dim \mathbb R^3 = 3$, any four vectors in $E$ would be linearly dependent, therefore 3. is true.

I am stuck on 2., as it asks about any 3 elements of $E$. And there are 10 choices for 3 elements out of 5. The only method i can think of is to reduce all these triplets in a matrix, and if the rank is 3 then they are linearly independent, but that will take much time.

How to reduce cases? Or to solve in one matrix altogether, if possible.

• Please use $\LaTeX$ – Aweygan Jan 26 '17 at 15:50
• Sorry, but i am writing this from my phone, and it is hanging if i try with $. – Shobhit Jan 26 '17 at 15:53 • @Aweygan : What is used here is MathJax, not LaTeX. LaTeX has zillions of features not found in MathJax. Whoever masters MathJax and thinks they know LaTeX will suffer a severe shock when they encounter actual LaTeX and find they don't know it. – Michael Hardy Jan 26 '17 at 16:24 1 Answer For$2)$, observe that $$\begin{pmatrix}1\\1\\1 \end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1 \end{pmatrix}$$ • What if there were 50 vectors in E, and we were asked if any 25 of them would be linearly independent? All i am saying is, other than keen observation, is there any other method? – Shobhit Jan 26 '17 at 15:56 • @Shobhit Well if the dimension of the vector space is$\leq 24$, then the answer is simple. If the dimension is$\geq 25$, then yes you would have to determine linear independence via checking each$25\$-element subset, or getting lucky and finding one which isn't. – Aweygan Jan 26 '17 at 16:01
• Ok, understood. Thank u. – Shobhit Jan 26 '17 at 16:06