Find a value r so that the vector v is in the span of a set of vectors

Question

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 $\mathbb{R}^N$

Answer 1

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.

Answer 2

You are correct, this problem was set up with the first coordinate nice. In general, if you have $n$ vectors in $\Bbb R^n$ with one unknown value, you can just form the determinant and choose the value that makes the determinant zero. This will be a linear equation, but will take a bunch of computation to get there. In your example, for $n=4$, there are $4!=24$ terms in the determinant. Six of them will involve $r$ and the other eighteen will not.

Answer 3

Your ad hoc solution is excellent. The general method consists in solving a system of linear equations, reducing it to row echelon form via Gaussian elimination. In your case you start with the matrix $$ \begin{bmatrix} 0&0&3&3\\ 0&3&3&r\\ -3&4&1&-10\\ 3&-3&5&14\\ \end{bmatrix} $$ and apply row elementary operations to get $$ \begin{bmatrix} 0&0&3&|&3\\ 0&3&3&|&r\\ 0&1&6&|&4\\ 3&-3&5&|&14\\ \end{bmatrix},\quad \begin{bmatrix} 0&0&3&|&3\\ 0&0&-15&|&r-12\\ 0&1&6&|&4\\ 3&-3&5&|&14\\ \end{bmatrix},\quad \begin{bmatrix} 0&0&-15&|&r-12\\ 0&0&3&|&3\\ 0&1&6&|&4\\ 3&-3&5&|&14\\ \end{bmatrix},\quad \begin{bmatrix} 0&0&0&|&r+3\\ 0&0&3&|&3\\ 0&1&6&|&4\\ 3&-3&5&|&14\\ \end{bmatrix}. $$ This shows that there is a solution only when $r+3 = 0$, that is, $r = -3$.