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## Answered question

2022-10-06

How to find linearly independent vectors?
There are three vectors:
${a}_{1}=\left(-1,1,0,x\right)\phantom{\rule{0ex}{0ex}}{a}_{2}=\left(2,-3,1,2\right)\phantom{\rule{0ex}{0ex}}{a}_{3}=\left(1,-2,1,-1\right)$
How can I find the parameter x so that these vectors are linearly independent? I'm not quite sure how to approach this problem since there are three vectors in ${\mathbb{R}}^{4}$, so I can't calculate the determinant of the matrix. Any help will be appreciated!

### Answer & Explanation

Baron Coffey

Beginner2022-10-07Added 5 answers

When you don't know what to do in a problem, go back to the definitions. It might not be the fastest or easiest way to do it, but it can usually give you some insight.
Going back to the definition of linearly independent. The vectors ${a}_{1},{a}_{2},{a}_{3}$ are linearly independent if ${c}_{1}{a}_{1}+{c}_{2}{a}_{2}+{c}_{3}{a}_{3}=\left(0,0,0,0\right)$ implies ${c}_{1},{c}_{2},{c}_{3}$ are zero, here ci are constants.
Just write down the expression, you will end up with a linear system on the variables ${c}_{1},{c}_{2},{c}_{3}$ (this system actually gives you a matrix to row-reduce, if you want it). Now, the x given in the problem will appear too. You just have to work with it a bit and you will get to the solution.

solvarmedw

Beginner2022-10-08Added 3 answers

It is easy to see that ${a}_{2}$ and ${a}_{3}$ are linearly independent, since the ratio of the components are different. For example $2/1\ne \left(-3\right)/\left(-2\right)$. Then we ask if ${a}_{1},{a}_{2},{a}_{3}$ are linearly independent. What does that mean? It means that if
$\alpha {a}_{1}+\beta {a}_{2}+\gamma {a}_{3}=0$
then $\alpha =\beta =\gamma =0$. You can verify by components:
$\begin{array}{rl}\alpha \left(-1\right)+\beta \left(2\right)+\gamma \left(1\right)& =0\\ \alpha \left(1\right)+\beta \left(-3\right)+\gamma \left(-2\right)& =0\\ \alpha \left(0\right)+\beta \left(1\right)+\gamma \left(1\right)& =0\\ \alpha \left(x\right)+\beta \left(2\right)+\gamma \left(-1\right)& =0\end{array}$
Since the first three equations do not contain x, we can verify if there is a non trivial solution. And indeed, if $\gamma =-\beta$ from the third equation, then the first two rows become:
$-\alpha +\beta =0\phantom{\rule{0ex}{0ex}}\alpha -\beta =0$
So $\alpha =\beta$, and you have a non trivial solution. The last of the equations for components is then
$\alpha \left(x+2+1\right)=0$
To have only the trivial solution $\alpha =0$ you must have $x\ne -3$. If $x=-3$ the equations are linearly dependent
${a}_{1}+{a}_{2}-{a}_{3}=0$

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