Linearly independent vectors with a missing coordinate

Question

For what values of $\lambda$, are $a_1=(\lambda, −1, −1)$, $a_2=(−1, \lambda, −1)$, $a_3=(−1, −1, \lambda) \in \Bbb R^3$, linearly independent?

From what I know, if $3$ vectors belong $\Bbb R^3$, then they are linearly independent so I don't know how to proceed with this problem. At first glance they seem linearly independent for any $\lambda$ except $-1$.

Answer 1

As you say, the determinant of the matrix is $\lambda^3 -3\lambda -2$. If the rows are to be linearly independent, we need nonzero determinant. So we set $\lambda^3 -3\lambda -2 = 0$ and figure out which values $\lambda$ cannot be.

Using synthetic division, we see $\lambda = 2$ is a root of that equation and that it factors to $(\lambda -2)(\lambda^2 +2\lambda +1) = (\lambda -2)(\lambda +1)^2 $.

Then setting $(\lambda -2)(\lambda +1)^2 = 0$, we see this only occurs when $\lambda =2, \lambda = -1$.

Then the rows are linearly independent when $\lambda \neq 2, \lambda \neq -1$.

Answer 2

Your determinant is

$$\lambda^3-3\lambda-2$$

$$=\lambda^3+1-3(\lambda+1)$$

$$=(\lambda+1)(\lambda^2-\lambda-2)$$

$$(\lambda+1)^2(\lambda-2).$$

thus, the three vectors are linearly independent if $\lambda\neq -1$ and $\lambda\neq 2$.