Linearly independent set.

Given $\displaystyle \vec{x_1} = (1, 2, -1,0),\vec{x_2} = (15 , 1, 0, 3), \vec{x_3} = (-2, 13, 3, -2)$. Find a vector $\displaystyle \vec{x_4}$ such that $\displaystyle A=\lbrace \vec{x_1}, \vec{x_2}, \vec{x_3},\vec{x_4}\rbrace$ is a linearly independent set.

Response:

Let $\displaystyle \vec{x_4}=(a_1,a_2,a_3,a_4)$ for arbitrary numbers $\displaystyle a_1, ..., a_4$.

Then for A to be linearly independent, $\displaystyle c_1\vec{x_1}+c_2\vec{x_2}+c_3\vec{x_3}+c_4\vec{x_4 }=\vec{0}$ must imply $\displaystyle c_1=c_2=c_3=c_4=0$.

Therefore we have the homogeneous linear system, in matrix form:

$\displaystyle \left( \begin{array}{cccc}

1&15&-2&a_1\\

2&1&13&a_2\\

-1&0&3&a_3\\

0&3&-2&a_4 \end{array}\right)

$

My question(s):

Do I really have to reduce it to reduced echelon form? It gets really messy!

Is there an easy way to solve this? (Easier than my approach?)

Generalized Cross Product

Cross product - Wikipedia, the free encyclopedia

You can look in the section on generalizations of the cross product into higher dimensions. This will give you a vector that is perpendicular to the subspace spanned by the first 3 vectors. If they are linearly independent, the collection of all 4 should be as well.

Alternatively, guess and check could be popular, you just gotta pick 4 numbers to give you a nonzero determinant that will show your vectors are linearly independent.

Otherwise, yeah, that is probably the best way to do it.