# Linear algebra dependence & matrices

• September 13th 2008, 06:21 PM
Linnus
Linear algebra dependence & matrices
Find the value(s) of h for which the vectors below are linearly dependent.

$\begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix}$

Thanks!
• September 13th 2008, 06:24 PM
ThePerfectHacker
Quote:

Originally Posted by Linnus
Find the value(s) of h for which the vectors below are linearly dependent.

$\begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix}$

Thanks!

$\det \left[ \begin{array}{ccc}1&3&-1\\-1&-5&5\\4&7&h \end{array}\right] = 0$

Does that make sense?
• September 14th 2008, 03:20 PM
Linnus
I'm confused on what exactly is linear independence. I googled it and this but all the explanations were confusing. Thanks!
• September 14th 2008, 05:54 PM
ThePerfectHacker
Quote:

$\begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix}$
These three vectors are linearly independent if we write:
$a \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}+b \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}+c \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix} = \bold{0} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
Then $a=b=c=0$.

This leads to the system of equations,
$a+3b-c = 0$
$-a-5b+5c=0$
$4a+7b+hc=0$

You want that this has a trivial solution i.e. $a=b=c=0$.

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 $h$ which give non-trivial solutions. Once you know that you can find what would give trivial solutions.