Finding real values for x that form dependet vectors

Question

I've got the following vectors

$v_{1}=(x,1,2)$, $v_{2}=(1,x,2)$ and $v_{3}=(1,2,x)$

So I my intial train of thought was to take compile the above vectors into a augmented matrix and take the determinant.

From theory it states that for coefficent matrices with a determinant which equals to zero is linearly dependet.

I thusly use this fact and take the determinant of the matrix.

From the determinant I would get create an expression interms of x that would equal zero.

However emplying this method I got wrong roots of x, so I might be missing something in my understanding of the problem.

Could somebody confirm if i'm on the right path or if I should reconsider?

Thank you.

Answer 1

I think your reasoning is right.

The determinant is $x^3-7x+6$, which gives $x\in\{1,2,-3\}$.

Because $$x^3-7x+6=x^3-x^2+x^2-x-6x+6=x^2(x-1)+x(x-1)-6(x-1)=$$ $$=(x-1)(x^2+x-6)=(x-1)(x^2-2x+3x-6)=(x-1)(x-2)(x+3).$$

Answer 2

This method is fine. Note that $$ \det\left(\begin{bmatrix}x&1&2\\1&x&2\\1&2&x\end{bmatrix}\right)= x(x^2-4)-(x-2)+2(2-x)=x^3-7x+6 $$ by expansion by minors. Now find the roots of the above.

Answer 3

In order to make these vectors dependent $v_{1}=(x,1,2)$, $v_{2}=(1,x,2)$,$v_{3}=(1,2,x)$, let $x=1$ and the first two vectors become identical.

Similarly for $ x=2$ you get The second and the third one identical.

For $x=-3$ you get $5v_1+7v_2+8v_3=0$