Find the value(s) of h for which the vectors below are linearly dependent.
$\displaystyle \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix} $
Thanks!
Find the value(s) of h for which the vectors below are linearly dependent.
$\displaystyle \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix} $
Thanks!
These three vectors are linearly independent if we write:$\displaystyle \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix} $
$\displaystyle a \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}+b \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}+c \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix} = \bold{0} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $
Then $\displaystyle a=b=c=0$.
This leads to the system of equations,
$\displaystyle a+3b-c = 0$
$\displaystyle -a-5b+5c=0$
$\displaystyle 4a+7b+hc=0$
You want that this has a trivial solution i.e. $\displaystyle a=b=c=0$.
To do that you need to compute the determinant of that system and make it equal to zero. And that will tell you all the values of $\displaystyle h$ which give non-trivial solutions. Once you know that you can find what would give trivial solutions.