Let S be = $\displaystyle

\begin{bmatrix}2 \\ 1 \\ 0 \\ 4 \\ \end{bmatrix} \begin{bmatrix}1 \\ 3 \\ 4 \\ -4 \\ \end{bmatrix} \begin{bmatrix}3 \\ 4 \\ 4 \\ 0 \\ \end{bmatrix} \begin{bmatrix}4 \\ -3 \\ -3 \\ -2 \\ \end{bmatrix}

$

and let W be the subspace spanned by S.

$\displaystyle U{\rightharpoonup}$ = $\displaystyle \begin{bmatrix}9 \\ 2 \\ 1 \\ 2 \\ \end{bmatrix} $

**Is $\displaystyle U{\rightharpoonup}$ in W? If so find a linear combination of the vectors in S Which adds up to U**
So i know that i am trying to find r1, r2, r3, r4, such that

$\displaystyle

r1*\begin{bmatrix}2 \\ 1 \\ 0 \\ 4 \\ \end{bmatrix} + r2*\begin{bmatrix}1 \\ 3 \\ 4 \\ -4 \\ \end{bmatrix} + r3*\begin{bmatrix}3 \\ 4 \\ 4 \\ 0 \\ \end{bmatrix} +r4*\begin{bmatrix}4 \\ -3 \\ -3 \\ -2 \\ \end{bmatrix} = \begin{bmatrix}9 \\ 2 \\ 1 \\ 2 \\ \end{bmatrix}

$

So i found the reduced row echelon form of the matrix and this is what i got

$\displaystyle \begin{bmatrix}1&0&1&0&|2 \\ 0&1&1&0&|1 \\ 0&0&0&1&|1 \\ 0&0&0&0&|0 \\ \end{bmatrix}

$

column 1 is x1 column 2 is x2 .... etc.

From this i can see that the 3rd column is free which shows that there are infinitely many solutions. So YES U is in W because the system has at least one solution.

so i paramaterized the solutions and set....

x3 = y

x1 + x3 = 2

x2 + x3 = 1

x4 = 1

so

x1 = 2-y

x2 = 1-y

x3 = y

x4 = 1

y can be any real # so... if y=1 then

r1 = 1

r2 = 0

r3 = 1

r4 = 1

This ^^ is one linear combination of the vectors in S that adds up to U

Did i do all this right? I just want to check. Thanks for looking it over and commenting on anything