2 years ago

Solve 765/14 and write your answer in terms of a remainder (please use this format: 59 R 7)

Mathematics

1 answer:

julsineya [31]2 years ago

8 0

The answer to this question is 54 R 9.

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False, counterexample below

Step-by-step explanation:

Denote the unit vectors of R^3 by $e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1)$ . Now consider $v_1=e_1, v_2=2e_1$ and $v_3=e_3$ . We have that $v_1, v_2,v_3 \in \mathbb{R}^3$ . Also, the vector $v_3$ is not a linear combination of $v_1, v_2$ because any linear combination of these two vectors will have third coordinate zero, but v_3 has third coordinate 1 so they can't be equal.

However, the set $\{v_1, v_2,v_3\}$ is not linearly independent, because $2v_1-v_2+0v_3=0$ is a non-trivial linear combination of these vectors that equals zero.