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Let v1; v2; and v3 be three nonzero vectors in R3. Suppose v2 is not a scalar multiple of either v1 or v3 and v3 is not a scalar multiple of either v1 or v2. Does it follow that every vector in R3 is is a linear combination of v1, v2, and v3?

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  1. 5 May, 00:03
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    Not necessarily

    Step-by-step explanation:

    Lets take v1 = (1,0,0), v2 = (1,1,0) and v3 = (0,1,0). Neither of the vectors are a multiple of the other, however they dont generate R³ because for example the vector (0,0,1) is not a linear combination of v1, v2 and v3.

    Not that, despite not being a multiple of v1 or v3, v2 is a linear combination of v1 and v3, because it is the sum of both of them. Therefore, the three vectors are linearly dependent and they cant generate a 3 dimensional vector subspace.
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