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How do you solve this problem?

Write each vector as a linear combination of the vectors in S if possible:

$S = \{(2,0,7),(2,4,5),(2,-12,13)\}$ $u = (-1,5,-6)$

I only got to the point where left hand side of the two equations has same value which evaluates to zero. Here's how my attempt look like:

$a(2,0,7) + b(2,4,5) + c(2,-12,13) = (-1,5,-6)$

$$2a + 2b + 2c = -1 \tag{1}$$ $$ 4b - 12c= 5 \tag{2}$$ $$7a + 5b + 13c= -6 \tag{3}$$ $$EQ(1) \times (-2) + EQ(2) = 4a + 16c = -7 \tag{4}$$

$$EQ(1)/2 \times 5 + EQ(3) = 2a + 8c = -7/2 \tag{5}$$

As you can see, I can't eliminate either term. The left side will evaluates to zero. Where did I do wrong? According to answer keys, I should get $a=-7/4, b= 5/4, c=0$.

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suppose you want to find a linear combination for the vector $(x_1,x_2,x_3)$ And $a,b$ and $c$ are the coefficients for vectors $(2,0,7),(2,4,5)$ and $(2,-12,13)$ respectively.

Then you get the following system of linear equations:




Write the augmented matrix:

\begin{array}{cc} 2 & 2 & 2 & x_1\\ 0 & 4 &-12 & x_2\\ 7 & 5 & 13 &x_3\\ \end{array}

subtract $\frac{7}{2}$ times the first equation from the last one to get

\begin{array}{cc} 2 & 2 & 2 & x_1\\ 0 & 4 &-12 & x_2\\ 0 & -2 & 6 &x_3-\frac{7}{2}x_1\\ \end{array}

Now add two times the third equation to the second one to get:

\begin{array}{cc} 2 & 2 & 2 & x_1\\ 0 & 0 &0 & x_2+2x_3-7x_1\\ 0 & -2 & 6 &x_3-\frac{7}{2}x_1\\ \end{array}.

So if a vector is a linear combination of these vectors: if the vector has entries $(x_1,x_2,x_3)$ then $x_2+2x_3-7x_1=0$

$5-12+7=0$ so there is no problem in that aspect. lets keep working on our matrix: lets add the third vector to the first:

\begin{array}{cc} 2 & 0 & 8 & x_1+x_3-\frac{7}{2}x_1\\ 0 & 0 &0 & x_2+2x_3-7x_1\\ 0 & -2 & 6 &x_3-\frac{7}{2}x_1\\ \end{array}.

From here we get the following system of equations:



Therefore fore any c there is a solution:

Lets plug in our numbers and find a solution for c= 0 so then $2a=-1-6+\frac{7}{2}$ so $a=\frac{-7}{4}$

and $-2b=-6+\frac{7}{2}$ so $b=\frac{5}{4}$

plugging this into the original equation we get $(-\frac{-7}{2},0,-\frac{49}{4})+(5/2,5,\frac{25}{4})=(-1,5,-6)$ as desired. However for any real value of c there is a solution.

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Thanks! But I don't get two things. One is - 5−12+7=0 where did you get those numbers? The other is 2a = 2a=−1−6+72 – Andy Mar 17 '13 at 12:24
Yes, for a vector to have a linear representation using the other three, the vector needs to have the property that $x_2+2x_3-7x_1=0$. This is because we got to the equation $0a+0b+0c=x_2+2x_3-7x_1$ which can only be satisfied when the right part is zero. I was doing 5-12+7 to check that this was in fact the case in vector (-1,5,-6). If this wasn't the case then it would be impossible. – Carry on Smiling Mar 17 '13 at 18:00

First, let me say that you need to be more careful with your work. Writing $EQ(1) + EQ(2) = 2a + 6b -10c = 4$ is a little sloppy. Rather, you should write what $EQ(1) + EQ(2) \implies 2a + 6b - 10c = 4$, or just say "Adding equation 1 and equation 2" gives...

Second, what you write is not quite correct, but rather equivalent to the equations you want. For example:

$$ EQ(1)\times(−2)+EQ(2) \implies -4a-16c=7 $$ which is almost the equation you wrote, but you're off my a negative sign.

With this in mind, you have not made any arithmetic mistakes. Equations (4) and (5) are the same, and they do not contradict each other. If you prefer substitution, solve for one of the variables in the second equation, and plug that variable into the others. This gives two equations and two unknowns. For example, solving for $b$ in the second equation we get $b = \frac{5+12c}{4} = \frac{5}{4} + 3c$. Thus:

$$ 2a + 2\left(\frac{5}{4} + 3c\right) + 2c = -1 \\ 7a + 5 \left(\frac{5}{4} + 3c \right) + 13c = -6 $$ has two equations and two unknowns. Again, I will let you fill in the details. I should point out, however, that substitution is pretty ugly as soon as we have more than two equations or more than two variables. In this case, I would suggest using Matrix (row) operations:

$$\left[\begin{array}{ccc|c} 2 & 2 & 2 & -1 \\ 0 & 4 & -12 & 5 \\ 7 & 5 & 13 & -6 \end{array}\right]$$

Finally, regardless of whatever method you use to solve the equation, you should note that you will get infinitely many solutions (your solution set will be a plane in $\Bbb R^3$), but this fine as all you need is at least one solution.

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Thanks! But I still can't get substitution part right. -4a-16c = 7, -2a-8c=-5/2. When I solve for a and plug in to the 2nd equation, all c terms cancelled each other out and evaluates to zero. I appreciate if you could show that part as well. Thanks! – Andy Mar 17 '13 at 12:58

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