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[tex]

A = \left[ {\begin{array}{*{20}c}

1 & 2 & 3 & 1 \\

2 & 4 & 8 & { - 2} \\

{ - 5} & { - 10} & { - 19} & 3 \\

\end{array}} \right]

[/tex]

[tex]

B = \left[ {\begin{array}{*{20}c}

1 & 2 & 0 & 7 \\

0 & 0 & 1 & { - 2} \\

0 & 0 & 0 & 0 \\

\end{array}} \right]

[/tex]

B is the row reduced echelon form of the matrix A.

a) What is the rank of A.

2.

b) What down a basis for the column space of A.

{(1,2,-5),(3,8,-19)}

c) Find the dimension of the row space of A.

2.

d) Are the rows of A linearly independent? Explain your answer.

No, rank(row space) > dimension(row space).

e) Do the vectors (1,2,-5), (3,8,-19) and (1,-2,3) span R^3?

The answer says something along the lines of "The 3 vectors are in the column space of A which has dim=2. This implies that the 3 vectors span a subspace with dim <= 2. So they do not span R^3."

I'm not completely about how to get to the answer. I know that dim(R^3) = 3 but I don't see why the 3 vectors span a subspace which has dimension less than or equal to 2. Further, I can't understand why it follows that they cannot span a subspace greater than 2.

Because I don't understand this and also due to the fact people seem to prefer seeing an attempt from the OP I will try to explain what I have tried. I wasn't really sure about this question from the beginning so I went back to the definitions.

A set of vectors T spans a subspace of R^m, S, if every vector in S can be written as a linear combination of T(I think:D). In this case the set S

*is*R^3 and T is the set of the 3 given vectors. So yes, I think I could just let v = (x,y,z) to be an arbitrary vector in R^3, equate coefficients, write the corresponding augmented matrix and deduce the answer. However, the way that the question is set out and the way that the answer is phrased suggests that I do not need to do that.

In any case, looking at the answer I see that there is a mention of the dimension of the column space of A, which is equal to the rank of A. Ok well at this point I'm just confusing myself and its getting late so I'm finding it hard to think. So could someone please provide me with some assistance.