Finding linear dependency in two different ways Find the value(s) of h for which the vectors are linearly dependent. Justify each answer:
$
    \begin{bmatrix}
     1       & -2 & 3 \\
     5       & -9 & h \\
     -3       & 6 & -9 
    \end{bmatrix}
$ 
The correct answer is all $h$, but I'm confused on how to arrive at this answer. At first, I looked at the 2nd and 3rd column and noticed the values in the 2nd column could be multiplied by $\frac{-3}{2}$ to get the values in the 3rd column. I concluded $h$ must be $\frac{27}{2}$.
I also tried to put the matrix in row-reduced echelon form to get:
$
    \begin{bmatrix}
     1       & 0 & 2h-27 \\
     0       & 1 & h-15 \\
     0       & 0 & 0 
    \end{bmatrix}
$ 
, so I see that there is a free variable in th 3rd column, so $h$ can be any value since the system is already linearly dependent. I was wondering what was wrong with the way I did it with my first method.

More ways to solve this (among others):


*

*You can simply look at the first and last rows of the matrix which are clearly dependent.

*Calculating the determinant
$\begin{align*}
\begin{vmatrix}
1&-2&3\\
5&-9&h\\
-3&6&-9\\
\end{vmatrix} &= 1\begin{vmatrix}-9&h\\6&-9\end{vmatrix}+
                2\begin{vmatrix}5&h\\-3&-9\end{vmatrix}+
                3\begin{vmatrix}5&-9\\-3&6\end{vmatrix}\\
              &=81-6h-90+6h+90-81\\
              &=0\end{align*}$
Your first solution simply showed one value of $h$ for which the vectors are dependent. Using the same method as your first one, you could have subtracted the second column from the first to get the third, giving you $h=14$, again just one possible value of $h$.
