[Math] Writing u as a linear combination of the vectors in S.

linear algebramatricesvector-spaces

Write vector
u = $$\left[\begin{array}{ccc|c}2 \\10 \\1\end{array}\right]$$

as a linear combination of the vectors in S. Use elementary row operations on an augmented matrix to find the necessary coefficients.

S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , v2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
v2\left[\begin{matrix}5\\4\\1\end{matrix}\right]
$ }. If it is not possible, explain why?


This is what i have so far:

S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , v2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
v3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
v4\left[\begin{matrix}2\\10\\1\end{matrix}\right]
$ }

$c1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , c2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
c4\left[\begin{matrix}2\\10\\1\end{matrix}\right]
=\left[\begin{matrix}0\\0\\0\end{matrix}\right]$

$c1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , c2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
c4\left[\begin{matrix}2\\10\\1\end{matrix}\right]$

$
\begin{bmatrix}
1 & 4 & 5 & 2\\
2 & 2 & 4 & 10\\
2 & 1 & 1 & 1\\
\end{bmatrix}
$

Now i don't know how to do this. Help will greatly be appreciated.

Thanks

Best Answer

$a\left[\begin{matrix}1\\2\\2\end{matrix}\right] + b\left[\begin{matrix}4\\2\\1\end{matrix}\right]+ c\left[\begin{matrix}5\\4\\1\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right]$.

Putting this into matrix form gives:

$\begin{bmatrix}1&4&5&2\\2&2&4&10\\2&1&1&1\end{bmatrix}\to\begin{bmatrix}1&4&5&2\\1&1&2&5\\2&1&1&1\end{bmatrix}\to\begin{bmatrix}1&4&5&2\\0&3&3&-3\\0&7&9&3\end{bmatrix}\\\to\begin{bmatrix}1&4&5&2\\0&1&1&-1\\ 0&7&9&3\end{bmatrix}\to\begin{bmatrix}1&0&1&6\\0&1&1&-1\\0&0&1&5\end{bmatrix}\to\begin{bmatrix}1&0&0&1\\0&1&0&-6\\0&0&1&5\end{bmatrix}$.

So, $a=1,b=-6,c=5.$

Plug the values back in to check if this is indeed correct:

$1\left[\begin{matrix}1\\2\\2\end{matrix}\right] -6\left[\begin{matrix}4\\2\\1\end{matrix}\right]+ 5\left[\begin{matrix}5\\4\\1\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right]$$\implies$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] + \left[\begin{matrix}-24\\-12\\-6\end{matrix}\right]+ \left[\begin{matrix}25\\20\\5\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right]$.

So, it must be correct.