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I have a question: Can the vector $v = (1,2)$ be expressed as a linear combination of $u_1 = (1,3)$ and $u_2 = (4,1)$?

What I have tried:

$a + 4b = 1$

$3a + b = 2$

$a = 1 - 4b$

$3(1 - 4b) +b = 2$

$3 - 12b + b = 2$

$3 -11b=2$

$3 -2 = 11b$

$1 = 11b$

$b = 1/11$

$a + 4(1/11) = 1$

$a =1 - 4/11$

$a = 7/11$

Therefore, $v = 7/11x + 1/11y$

That's what I got but I know it's wrong because it doesn't seem right! Any help about how to go about doing this will be much appreciated!

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You know it's wrong because "it doesn't seem right"? Well, that's not a very mathematical argument, is it? And BTW: it is correct. – DonAntonio Feb 11 '14 at 16:45

Your answer is indeed correct. You can (and should) always check your solution directly:

$$\frac{7}{11}(1,3) + \frac{1}{11}(4,1) = (\frac{7}{11},\frac{21}{11})+(\frac{4}{11},\frac{1}{11}) = (\frac{11}{11},\frac{22}{11}) = (1,2)$$

as desired.

It's generally true that as long as $u_2$ isn't a scalar multiple of $u_1$, you can find a unique linear combination of $u_1$ and $u_2$ to give any desired $v$. That's because the condition is exactly what it takes for $\{u_1,u_2\}$ to be a basis for $\mathbb{R}^2$.

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It's correct. You should try with row elimination next as it is much faster, and can be done in fewer steps.

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