Finding a basis in ImT using Gaussian Elimination

[JoshMaths]

Homework Statement

$$
\begin{pmatrix}
-1&3&0\\
2&0&-1\\
0&-6&1
\end{pmatrix}
$$

Finding the ImT basis of this

The Attempt at a Solution

I got it down to

$$
\begin{pmatrix}
1&0&-1/2\\
0&1&1/6\\
0&0&1
\end{pmatrix}
$$

I know that by the principle of having pivots as the only non-zero entities in their respective columns this makes that column one of the basis vectors. So answer is

[-1,2,0] [3,0,-6]

What i don't understand is why (In the r-echelon form) i cannot subtract -1/2(Row 3) from Row 1 and Subtract 1/6(Row 3) from Row 2 to give the Gauss-Jordan form or Identity form which would imply that the entire first matrix was a basis for itself right? Meaning the basis contains three vectors instead of the actual two is contains in the correct answer.

Thanks, and i hope the question has come out clearly, just say if clarification is needed.

Josh

 

[genericusrnme]

What is 'ImT basis'?
The image of some basis under T?

 

[JoshMaths]

No it's just the basis over T i guess. The generic basis that spans T(v) where v is an arbitrary vector and the matrix for T is the above.