How many columns of this matrix are linearly independent? \begin{matrix}
    1 & 2 & 0 & 1 \\
    2 & 4 & 1 & 4 \\
    3 & 6 & 3 & 9 \\
    \end{matrix}
I have tried to transpose it and then reduce it by row echelon form and i get zeros on the last two rows. But i can't grasp if i should be doing that or doing it another way.
 A: You are after the rank of that matrix. Every $3\times3$ submatrix has $0$ determinant, but there are $2\times2$ submatrices whose determinant is not $0$. So, the rank is $2$. So, any three columns are linearly dependent, but there are pairs of columns which are linearly independent (the second and the third column, for instance).
A: The row rank of a matrix A, r(A), is the dimension of the subspace of $F^k$ spanned by rows of A. 
For every two matrices A and B, if A~B then r(A)=r(B).
For every matrix A, $r(A) = r(A^T)$.Check Here
so by the last statement, you don't really need to Transpose the matrix
Basically, you just need to row reduce the matrix to its row echelon form then these set of vectors(all not all-zero rows) will be linearly independent. Cause each row which became all-zero were representable via a linear combination of the first and second row.
A: If you look at your matrix 
$$ A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 &4 \\ 3 & 6 & 3 & 9 \end{bmatrix} $$
The rank can be no more than $3$. Do the following
Subtract $2 R_{1} $ from $R_{2}$
$$ R_{2} - 2 R_{1} =  \begin{bmatrix} 2 & 4 & 1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 4 & 0 & 2\end{bmatrix} = \begin{bmatrix} 0 & 0& 1  & 2\end{bmatrix} $$
$$  A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 3 & 6 & 3 & 9 \end{bmatrix} $$ 
Now we need to zero the third row. So subtract $3 R_{1} $ from $R_{3}$ 
$$ R_{3} - 3R_{1} =  \begin{bmatrix} 3 & 6 & 3 & 9 \end{bmatrix} - \begin{bmatrix} 3 & 6 & 0 & 3\end{bmatrix} = \begin{bmatrix} 0 & 0& 3  & 6\end{bmatrix} $$
$$  A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 0 & 0 & 3 & 6 \end{bmatrix} $$
Now we need to zero that $3$ in row $3$ . So we subtract $3R_{2}$ from $R_{3}$ 
$$ R_{3} - 3R_{2} =  \begin{bmatrix} 0 & 0 & 3 & 6 \end{bmatrix} - \begin{bmatrix} 0 & 0 & 3 & 6\end{bmatrix} = \begin{bmatrix} 0 & 0& 0  & 0\end{bmatrix} $$
You're left with this matrix
$$  A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 &2 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
There are only $2$ non-zero rows and they aren't multiples of eachother. The rank is $2$
If you want to look at column-wise then $2C_{1} = C_{2}$ and $C_{4} = C_{1}+2C_{3}$.
