Linear Algebra - Can vector $v$ be expressed as a linear combination of $u_1$ and $u_2$ 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!

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 $\mathbb{R}^2$.
