# Thread: Checking for value that makes vectors linearly dependent

1. ## Checking for value that makes vectors linearly dependent

Problem on the past qualifier asks to find the value of h for which the vectors are linearly dependent:

$\displaystyle v_1=\begin{pmatrix} 1 \\ -1 \\ -3 \end{pmatrix},v_2=\begin{pmatrix} -5 \\ 7 \\ 8 \end{pmatrix},v_3=\begin{pmatrix} 1 \\ 1 \\ h \end{pmatrix}$

For vectors to be linearly dependent they need to be products of each other multiplied by some coefficient. By mental math I found that with $\displaystyle 6*v_1 + v_2 = v_3$ vectors are linearly dependent if $\displaystyle h = -10$.

Would someone please comment on the correctness and point me to the formulaic way to find such solutions?

2. ## Re: Checking for value that makes vectors linearly dependent

I would use:

$\displaystyle A\begin{pmatrix} 1 \\ -1 \\ -3 \end{pmatrix}+B\begin{pmatrix} -5 \\ 7 \\ 8 \end{pmatrix}+C\begin{pmatrix} 1 \\ 1 \\ h \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

Where $\displaystyle A,B,C\ne0$

Giving the system:

(1) $\displaystyle A-5B+C=0$

(2) $\displaystyle -A+7B+C=0$

(3) $\displaystyle -3A+8B+hC=0$

Adding (1) and (2) we find:

$\displaystyle B=-C$

Substituting for B into (1) we find:

$\displaystyle A=-6C$

Substituting for A and B into (3) we find:

$\displaystyle C(h+10)=0\:\therefore\:h=-10$

3. ## Re: Checking for value that makes vectors linearly dependent

Another way: $\displaystyle v_1,v_2, v_3\;\textrm{l.d.}\Leftrightarrow \textrm{rank}\;[v_1,v_2,v_3]<3\Leftrightarrow \det\;[v_1,v_2,v_3]=0\Leftrightarrow h=-10$