-1
$\begingroup$

Suppose (x;y; z) is a linear combination of (2;3;1) and (1;2;3). What determinant is zero? What equation does this give for the plane of all combinations?

Can someone give me a hint or show me how can I solve this, please?

EDIT

I used the determinant

row1 = x y z row2 = 3 2 1 row3 = 1 2 3

and I reached the equation 4x-8y+4z

I don't know if it's correct or if it's supposed to equal zero and how....

$\endgroup$

closed as unclear what you're asking by Thomas, Casteels, YoTengoUnLCD, user147263, Michael Medvinsky Dec 11 '15 at 0:24

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ @Justpassingby gave a pretty clear hint and partial answer. Isn't it clear that the determinant is "supposed to equal zero"? If you do not understand why, you should not be doing these kinds of problems. It looks like you need to review the ideas leading up to your current topic. $\endgroup$ – Rory Daulton Dec 11 '15 at 11:35
2
$\begingroup$

The condition that $(x;y;z)$ be linearly dependent on the two given vectors is equivalent to

$$\left| \begin{matrix} 1&2&3 \\ 2&3&1 \\ x&y&z \end{matrix} \right|=0.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.