Find the value r so that,
$$v = \begin{pmatrix} 3 \\r \\-10\\14 \end{pmatrix}$$
is in the set,
$$ S= \text{span}\left(\begin{pmatrix} 3\\3\\1\\5
\end{pmatrix}, \begin{pmatrix} 0\\3\\4\\-3 \end{pmatrix}, \begin{pmatrix} 0\\0\\-3\\3 \end{pmatrix} \right) $$
What I've tried is this…, brute force and intuition. Since we're looking for a linear combination of the 3 vectors in the set S, we know that we can only use a multiple of 1 from the first vector, call this $c_1=1$. We still need $c_2$ and $c_3$ (combinations of the second vector and third vector, respectively). One can see that if we let $c_2=-2$, and $c_3=1$, we can get the bottom two entries of $v$. Hence, the only value that $r$ can be is $-3$ , so $r=-3$.
This is great…, but how about when the system is really complicated and trial and error won't suffice, how would I be able to calculate an arbitrary $r$ that is in a vector $\in$$\mathbb{R}^N$. With a set of vectors that span $\mathbb{R}^N$
Best Answer
You can use your row reduction skills to solve this. We have 3 equations and 3 unknowns, where the unknowns are $c_1$, $c_2$ and $c_3$.
We have: $3c_1+0c_2+0c_3=3,1c_1+4c_2-3c_3=-10,5c_1-3c_2+3c_3=14$. This suggests the augmented matrix:
$$ \begin{pmatrix} 3&0&0&3\\1&4&-3&-10\\5&-3&3&14 \end{pmatrix} $$
When row reduced, this gives
$$ \begin{pmatrix} 1&0&0&1\\0&1&0&-2\\0&0&1&1 \end{pmatrix} $$
Which means $c_1=1,c_2=-2,c_3=1$.
Then, taking $3c_1+3c_2+0c_3=-3=r$.
This method works in general.