Please help me find out if the colums of $A$ form a basis of $\mathbb{R}^5$

Let $$A=\begin{pmatrix}3&1&4&1&5&9\\2&6&5&3&5&8\\9&7&9&3&2&3\\8&4&6&2&6&4\\3&3&8&3&2&7\end{pmatrix}.$$ Do the columns of $A$ form a basis of $\mathbb{R}^5$?

This is the rref.

$$A=\begin{pmatrix}1&0&0&0&0&-23/11\\0&1&0&0&0&4/3\\0&0&1&0&0&106/33\\0&0&0&1&0&-214/33\\0&0&0&0&1&50/33\end{pmatrix}.$$

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Uhh, digits of $\pi$. –  user1551 Mar 25 '13 at 22:15
Have you tried row reducing this matrix? What do you get? –  Jim Mar 25 '13 at 22:20
Yes, I did rref and got –  Walter Mar 25 '13 at 22:58
So the RREF has five linearly independent rows, and in turn the row rank of $A$ is $5$. Yet the row rank is always equal to the column rank. Hence $A$ has five linearly independent columns. –  user1551 Mar 25 '13 at 23:36

$A$ has six columns, but $\dim\mathbb{R}^5=5$. So the columns of $A$ certainly do not form a basis of $\mathbb{R}^5$. So I assume you meant to ask whether the columns of $A$ span $\mathbb{R}^5$. This is equivalent to showing that the rank of $A$ is $5$. So, you can use elementary row/column operations to transform $A$ to a row/column reduced echelon form, and check how many nonzero rows/columns are there.
It has $6$ columns, each of dimension $5$. A space has dimension $5$ (by definition) if and only if it has a basis with $5$ basis vectors. A basis just corresponds to a coordinate system (which is probably oblique and the lengths are not necessarily equally measured on the axes).
There's a basic theorem of vector spaces, that if any basis has $n$ elements, then all bases have $n$ elements. In consequence, no $6$ vectors can form a basis for the $5$ dimensional space.
There are six columns, so, no, they don't form a basis. However, any $5$ of them do form a basis. The culomns form a basis iff the elimination can give you the identity matrix. –  Berci Mar 25 '13 at 23:05