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Determine which sets in Exercises $1-8$ are bases for $\mathbb{R}^{3}$ . Of the sets that are not bases, determine which ones are linearly independent and which ones span $\mathbb{R}^{3}$ . Justify your answers.$$\left[\begin{array}{r}{1} \\ {-3} \\ {0}\end{array}\right],\left[\begin{array}{r}{-2} \\ {9} \\ {0}\end{array}\right],\left[\begin{array}{l}{0} \\ {0} \\ {0}\end{array}\right],\left[\begin{array}{r}{0} \\ {-3} \\ {5}\end{array}\right]$$

the given set of vectors spans $R^{3}$

Calculus 3

Chapter 4

Vector Spaces

Section 3

Linearly Independent Sets; Bases

Vectors

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in this exercise, a set of four different vectors are provided, and our goal here is to determine if these vectors form a basis for our three. Let's start off by making a knob survey shin that if we form a set the one v to V three N V four. That this set is automatically Lee nearly dependent and it's linearly dependent for really two reasons. But let's go with the big glaring reason This is linearly dependent. Since V three is equal, toothy zero vector we know any set that contains a The zero vector is automatically a linearly dependent set. It's also linearly dependent because we have four vectors with three entries and we also know a set with more vectors than entries per vector must also be dependent. So now that we know it's literally dependent, it also follows that if we call it s, that's the set s does not form ah, basis four r three And the reason for this is our basis must be linearly independent, and so obtaining linear dependence tells us we do not form a basis in this case. Now, another question we might have for this particular set is sure that it vectors are linearly dependent. They don't form a basis. But would they spend our three? Well, one waited to determine that would be to let a be a matrix formed from V one V two V three end before the vectors in question. Now, if we ro reduce this matrix, we find that it's row equivalent to 100 negative to 30 The zero vector remains unchanged. B three and we have zero negative. Three and five. So this is an echelon form. When we wrote reduce analyzing this echelon form, we see a pivot here, here and here so we can conclude that since A has a pivot in every row, we now know that s which consists of these vectors. Spans are three. Or in other words, if we spend s that contained the four vectors, we will have our three. So the set is linearly dependent does not form a basis. But that does span

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