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So I have problem number 4:

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So what I tried doing is that I set the vector from a) equal to c1*(first vector from S) and c2*(second vector from S), and from that I got the 4x3 matrix:

[6,4,-42;-7,6,13;8,-4,-112;6,1,-60]

And this is impossible to solve since this is basically 4 equations with 3 unknowns. Does this lead to the conclusion that a b c and d cannot be written as a linear combination of S ?

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Let $A= \begin{bmatrix} 6 & 4\\ -7 & 6\\ 8 & -4\\ 6 & 1 \end{bmatrix} $ be a matrix with column vectors from $S$. Let $\textbf{b}$ be the given vectors which you want to solve for. To find the possible coefficients of these two vectors, you want to solve a system $\textbf{A}x=\textbf{b}$. In augmented matrix form, for (a), this is $ \begin{bmatrix} 6 & 4 & -42\\ -7 & 6 & 113\\ 8 & -4 & -112\\ 6 & 1 & -60 \end{bmatrix}. $ Row reducing gives $ \begin{bmatrix} 1 & 0 & -11\\ 0 & 1 & 6\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix} $ so $x=\begin{bmatrix}-11 \\ 6\end{bmatrix}$, and you see

$ -11\begin{bmatrix} 6\\ -7\\ 8\\ 6\\ \end{bmatrix} +6 \begin{bmatrix} 4\\ 6\\ -4\\ 1\\ \end{bmatrix} = \begin{bmatrix} -42\\ 113\\ -112\\ -60\\ \end{bmatrix}.$ The others follow similarly. If you can find some solution vector $x$, then you can write the given vector $\textbf{b}$ as a linear combination of the vectors in $S$, otherwise you cannot.

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    @Virtuoso, yes, I seem to have gotten the same results as you. There is a pivot in the last column $f$or the augmented matrices for b) c) and d).2011-06-25
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I don't understand what you've been doing. Let's call $u$ and $v$ the vectors from $S$ and $w$ the one from (a). What do you want to know is if there exists real numbers $x$ and $y$ such that

$ w = xu + y v \ . $

Right?

But this is a system of linear equations, with four equations (right) and just two unknowns ($x$ and $y$). This one:

$ \begin{pmatrix} -42 \\ 113 \\ -112 \\ -60 \end{pmatrix} = x \begin{pmatrix} 6 \\ -7 \\ 8 \\ 6 \end{pmatrix} + y \begin{pmatrix} 4 \\ 6 \\ -4 \\ 1 \end{pmatrix} $

Which you can write as

$ \begin{pmatrix} 6 & 4 & \vert & -42 \\ -7 & 6 &\vert & 113 \\ 8 & -4 &\vert & -112 \\ 6 & 1 & \vert & -60 \end{pmatrix} $

If I'm not wrong, this time this system has no solution at all (and so, $w$ is NOT a linear combination of $u$ and $v$), but you don't have to think that, just because a system of linear equations has more unknowns than equations this necessarily means that it has no solutions. For instance, this one

$ \begin{align} x + y &=& 1 \\ x + y &=& 1 \\ x + y &=& 1 \end{align} $

Has two unknowns and three equations, but an infinite number of solutions. "Oh, but it's always the same equation, so it doesn't count". -Is that what you're thinking? Well, try to solve this one:

$ \begin{align} x + y &=& 1 \\ x - y &=& 0 \\ 2x &=& 1 \end{align} $

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    well you might want to take a look at the other answerer... supposedly he found a solution (yunone)2011-06-25