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3 Determine whether the setS = {(5,6,5) , (2,1,-5), (0, -4,1)}spans R3Determine whether S is basis for R3_S = {(0,0,0), (1,5,6) , (6,2,1)}...

Question

3 Determine whether the setS = {(5,6,5) , (2,1,-5), (0, -4,1)}spans R3Determine whether S is basis for R3_S = {(0,0,0), (1,5,6) , (6,2,1)}

3 Determine whether the set S = {(5,6,5) , (2,1,-5), (0, -4,1)} spans R3 Determine whether S is basis for R3_ S = {(0,0,0), (1,5,6) , (6,2,1)}



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Determine whether the set $S$ spans $R^{3} .$ If the set does not span $R^{3},$ then give a geometric description of the subspace that it does span. $S=\{(1,0,3),(2,0,-1),(4,0,5),(2,0,6)\}$

In this example, a set of vectors have has been provided. A miracle here is to determine if they form a basis for the space are three. First, let's look at the set V one, V two, and notice that we have a set of exactly two vectors. In this case, we can determine linear independence as follows. Let's determine if they could potentially be multiples of each other. In other words, could be multiply V one by some constant and obtained V to that constant being non zero. Well, it turns out that there is a major challenge to this. Our entry zero here and our entry five here in the corresponding vector V two tells us that no matter what value we put in from the V one toe, multiply the zero. We will never get a value five so automatically for the set V one V two. The set is Lee nearly independent, since V one and V two are not multiples of each other. If they were multiples of each other, we would say that the layer linearly dependent. So now we know we have linear independence. If these two vectors span are three, then they would also form the basis to resolve spanning. Let's start by setting a matrix, Let the Matrix a be given by V one and V two. Then this matrix, if we write it out, is negative. 230 six, negative one and five and without riddled row reduction, we know that a cannot have a pivot in every row. The reason for this being is if we have a pivot here, then in the best case, we have to pivots another pivot position here in an echelon form. But we could never obtain 1/3 pivot unless we had 1/3 column. So since they cannot have a pivot in every row, week now can conclude based on this statement, that the columns of A, which are V one and V two do not span, are three. So because we cannot have a pivot in every row, the columns do not span are three itself. So let's conclude this example. We know that the set is linearly independent but does not span. So our conclusion here is that the set of vectors V one and V two is not a basis. Four are to excuse me are three. If it spend that, it would have been a basis


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