Values of m so that the vectors are Linear dependent

Question

a) $u = (m, 1, m), v = (1, m, 1)$

b) $u = (m, 1, m + 1), v = (0, 1, m + 1), w = (0, m, 2m)$

I want to be sure that I got this questions right, can anyone help me if I got something wrong?

For a):

I got $m = \pm 1$

For b):

I got that $m = 0$

Answer 1

For (a) your answer is correct, in fact $m = ±1$ are the only possibilities such that the vectors are linearly dependent.

Check your work for (b), plugging $m=0$ in $u,v$ shows that $u=v$, so they are linearly dependent.

Here's how you work out all values of m. Assume that $m≠0$. $u,v,w$ are linearly independent iff there exist constants $a_1,a_2,a_3$, not all zero, such that $a_1 u + a_2 v + a_3 w = 0$. This set up a linear system of equations with three equations and three unknowns by matching coordinates in the vectors.

\begin{cases} ma_1 = 0 &\\ a_1+a_2+ma_3 = 0 &\\ (m+1)a_1 + (m+1)a_2 + 2m a_3 = 0 \end{cases}

We know that $m≠0$, so the first equation implies $a_1 = 0$. Keep reducing the system you'll find

\begin{cases} (m-1)a_3=0\\ a_2 + a_3 = 0 \end{cases}

The system only has non trivial solutions iff $m=1$, therefore $m=1$.

Answer 2

I can't see any other values that work for m other than $\pm 1$ so the first answer seems right,.. but the second, if you do some row operations to reduce matrix of coefficients or even if you just tried plugging some numbers into m, you will realize that any value for m would be sufficient for writing any of the vectors as a linear combination of the other two, so the answer is actually m = i: $\forall $ $i \in \mathbb{R}$