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I am trying to write \begin{bmatrix} 2 \\[0.3em] 3 \end{bmatrix} as a linear combination of

\begin{bmatrix} 1 \\[0.3em] -1 \end{bmatrix}

\begin{bmatrix} 6 \\[0.3em] 7 \end{bmatrix}

I concluded that I couldn't. This is wrong but I don't see how I can combine these, what steps do I need to take to do this?

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Check this thread out: math.stackexchange.com/questions/521503/…. It has similar stuff. –  Sujaan Kunalan Oct 22 '13 at 17:47
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5 Answers

If it wants to be true, you should find $a,b\in\mathbb R$ such that $$a(1,-1)+b(6,7)=(2,3)$$ Equivalently, you should check if the following system has any solution or not: $$a+6b=2,~~-a+7b=3$$ Note that $\begin{vmatrix} 1 & 6 \\ -1 & 7 \\ \end{vmatrix}=7+6=13\neq0$

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Nice use of the determinant +1:) –  Sami Ben Romdhane Oct 22 '13 at 17:55
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I'll assume that these are vectors in $\mathbb{R}^2$. Then what you want to do is find $a$ and $b$ so that $a\begin{bmatrix} 1\\ -1\end{bmatrix} + b\begin{bmatrix}6\\ 7\end{bmatrix} = \begin{bmatrix}2\\ 3\end{bmatrix}$ by solving the system of equations

$$\left\{\begin{aligned}a+6b &=2 \\ -a+7b &= 3\end{aligned}\right.$$

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Note that

$\begin{bmatrix} 2 \\ 3 \end{bmatrix} = a \begin{bmatrix} 1 \\ -1 \end{bmatrix} + b \begin{bmatrix} 6 \\ 7 \end{bmatrix} = \begin{bmatrix} 1 & 6 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}$,

and now solve a simple system of eqautions.

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If you write $\begin{bmatrix} 2 \\[0.3em] 3 \end{bmatrix}=a\begin{bmatrix} 1 \\[0.3em] -1 \end{bmatrix}+b\begin{bmatrix} 6 \\[0.3em] 7 \end{bmatrix}$ you get $2=a+6b,3=-a+7b$

Adding the two gives $5=13b$ and we find $b=\frac5{13},a=-\frac {4}{13}$

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How was I suppose to know to use algebra like that? I tried to make it into a reduced row echelon form because this is in a linear algebra class. I spent about 45 minutes on it and I couldn't do it. How do I know when to give up on RRE? –  Paul the Pirate Oct 22 '13 at 17:39
    
If you show the work on RRE we can take a look. It basically does the same thing and should be able to solve the problem –  Ross Millikan Oct 22 '13 at 17:45
    
I just stayed that 1,-1 can't be scaled in any way to make it equal since all the other numbers are positive. Logically this makes sense right? Since I am trying to find something to multiply 1,-1 by. –  Paul the Pirate Oct 22 '13 at 17:47
    
What are you trying to make it equal to? You are correct that it will not have all positive entries, but that is not the goal. –  Ross Millikan Oct 22 '13 at 17:51
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But you can multiply 6,7 by something as well. Without that you don't have enough freedom. –  Ross Millikan Oct 22 '13 at 18:01
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Set

$\begin{bmatrix} 2 \\ 3 \end{bmatrix} = a\begin{bmatrix} 1 \\ -1 \end{bmatrix} + b\begin{bmatrix} 6 \\ 7 \end{bmatrix}; \tag{1}$

then

$a + 6b = 2, \tag{2}$

and

$-a + 7b = 3. \tag{3}$

Adding these equations yields

$13b = 5, \tag{4}$

or

$b = \frac{5}{13}. \tag{5}$

Using this value of b in (2) we obtain

$a = 2 - \frac{30}{13} = -\frac{4}{13}. \tag{6}$

It is easily checked that these values of $a$ and $b$ satisfy (1), (2), and (3).

The point is, of course, that (1) is really just a simple linear system in the two variables $a$ and $b$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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