# Find the value(s) of h for which the vectors are linearly dependent. Justify eac

Find the value(s) of h for which the vectors are linearly dependent. Justify each answer.
$\left[\begin{array}{c}1\\ 5\\ -3\end{array}\right],\left[\begin{array}{c}-2\\ -9\\ 6\end{array}\right],\left[\begin{array}{c}3\\ h\\ -9\end{array}\right]$
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Step 1
is linearly dependent

Step 2

$A=\left[\begin{array}{ccc}1& -2& 3\\ 5& -9& h\\ -3& 6& -9\end{array}\right]$
$x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]$
Step 3

Step 4
From here we see that for any value of h, the possible solutions of ${A}_{x}=0$ are
$\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}\left(27-2h\right){x}_{3}\\ \left(15-h\right){x}_{3}\\ {x}_{3}\end{array}\right]={x}_{3}\left[\begin{array}{c}27-2h\\ 15-h\\ 1\end{array}\right]$
Since the last coordinate can be anything and is independent of h , we see that regardless of the value of h, this set of vectors is always linearly dependent.
Result:
The vectors are linearly dependent for all h