For what values of $h$ is $\mathbf{v_3}$ in the span of $\{\mathbf{v_1},\mathbf{v_2}\}$ ? For what values of $h$ is $\{\mathbf{v_1,v_2,v_3}\}$ linearly dependent?
$\mathbf{v_1}=\begin{bmatrix}1\\
-5\\
-3\end{bmatrix},\mathbf{v_2}=\begin{bmatrix}-2\\
10\\
6\end{bmatrix},\mathbf{v_3}=\begin{bmatrix}2\\
-10\\
h\end{bmatrix}$
My reasoning for the first part is as follows:
If $\mathbf{v_3}$ is in the span of $\{\mathbf{v_1,v_2}\}$ then it is a linear combination of $\mathbf{v_1,v_2}$. Writing an augmented matrix we have $$\begin{bmatrix}1&-2&2\\
-5&10&-10\\
-3&6&h\end{bmatrix}\sim\begin{bmatrix}1&-2&2\\
0&0&h+6\\
0&0&0\end{bmatrix}$$
Then this system is consistent only for $h=-6$, so $\mathbf{v_3}$ is in the span of $\{\mathbf{v_1,v_2}\}$ for $h=-6$.
For the second part, we just observe that $\mathbf{v_1}$ is a scalar multiple of $\mathbf{v_2}$ and therefore $\{\mathbf{v_1,v_2,v_3}\}$ must be linearly dependent for all values of $h$.
Is my reasoning here correct? I'm fairly confident with what I have but it seems to contradict the answer given in the text. Just looking for solution verification here. Thanks!
Best Answer
For part one, note that $v_1$ and $v_2$ are linearly dependent. They are multiples of each other. Thus we easily get that $v_3$ must be a multiple of $v_1$, and thus $h=-6$. Thus there was a typo.
For part two, let's use an alternate method and see if the answers agree. Compute the determinant. Get $10h+60+2(-5h-30)+2(0)=0$. This indicates that any $h$ gives you a linearly dependent set. Of course, this is obvious, by the first observation.