# 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.

Since the rank of a matrix and its transpose are equal, examining the rows of the matrix is a good way to go here. The third row is obviously a multiple of the first, so no matter what $h$ is, the rank of the matrix is at most 2, so its columns are also linearly dependent for any value of $h$.
\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 $h$.