vector space R³ 
Determining the three vectors are linearly dependent or independent
  $$u = (1, 2, -3), v = (3, 1, -2), w =(5, -5, 6). $$

Is looking through determinant enough to check whether linearly dependent or independent?

I know that determinant of the matrix is zero. So, Can I say it is linearly independent because of that determinant is zero ? To find the relation between $u$, $v$, and $w$ we look for constants a, b, and c. When I apply the echelon form I get  $\begin{bmatrix}1 & 0 & -4\\0& 1 & 3\\0 &0 &0\end{bmatrix}$. Up to here I understand, but I'm stuck here

how can we select $c = 1$ ? How come $b = -3c$ and $a = -3b - 5c $ ? Written $u$, $v$, and $w$ are linearly independent. But, the determinant is zero? I'm confused.

For an $n\times n$ matrix $A$ the following statements are equivalent:


*

*$A$ is invertible

*the columns of $A$ are linearly independent

*the rows of $A$ are linearly independent

*the rank of $A$ is $n$

*the determinant of $A$ is nonzero


So the answer to your question is yes, we need only check the determinant of 
$$
A=
\begin{bmatrix}
1&3&5\\
2&1&-5\\
-3&-2&6
\end{bmatrix}
$$
is zero to prove that the columns of $A$ are linearly dependent. Here,
$$
\det\begin{bmatrix}
1&3&5\\
2&1&-5\\
-3&-2&6
\end{bmatrix}
=0
$$
so the columns are indeed linearly dependent.
Alternatively we can use the fact that
$$
\DeclareMathOperator{rref}{rref}\rref
\begin{bmatrix}
1&3&5\\
2&1&-5\\
-3&-2&6
\end{bmatrix}
=
\begin{bmatrix}
1&0&-4\\
0&1&3\\
0&0&0
\end{bmatrix}
$$
to deduce that $\DeclareMathOperator{rank}{rank}\rank(A)=2<3$ so the columns of $A$ must be linearly dependent.
