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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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2 Answers 2

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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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Well I guess my issue is that my calculator won't perform row reducing for me. I have a TI 83+ and when I request an rref([A]) operation on the "augemented matrix form" matrix it says "INVALID DIM" –  Virtuoso Jun 25 '11 at 3:40
    
@Virtuoso, that's odd, if anything, matrices of this dimension should be easy enough to reduce by hand. But if that's too time consuming, you can always put it into this online reduction calculator to save time. –  yunone Jun 25 '11 at 3:43
    
Just for verification if you have time, b) c) and d) are all inconsistent, thus have no solution... correct? –  Virtuoso Jun 25 '11 at 15:06
    
@Virtuoso, yes, I seem to have gotten the same results as you. There is a pivot in the last column for the augmented matrices for b) c) and d). –  yunone Jun 25 '11 at 17:28

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) –  Virtuoso Jun 25 '11 at 3:43

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