Simplify: (a+2b+3c, 4a+5b+6c, 7a+8b+9c)

[courteous]

Homework Statement

"Simplify: (a+2b+3c, 4a+5b+6c, 7a+8b+9c)."
All of these are vectors (a=\vec{a}, etc.)

Homework Equations

I know that set of vectors \vec{x}_1+\cdots+\vec{x}_n \in V is linearly independent
if linear combination \alpha_1\vec{x}_1+\cdots+\alpha_n\vec{x}_n equals \vec{0}
iff all scalars \alpha_1\,\cdots,\alpha_n equal 0.

The Attempt at a Solution

(I don't understand it, although it's (partially) in my notebook)

(a+2b+3c, 4a+5b+6c, 7a+8b+9c) =
= (a,4a,7a) + (a,4a,8b) + (a,4a,9c) +
+ (a,5b,7a) + (a,5b,8b) + (a,5b,7a) +
+ (a,6c,8b) + ____0___ + ___0____ +
+ (2b,4a,9c) + (2b,6c,7a) +
+ (3c,4a,8b) + (3c,5b,7a) + =
= 48(a,b,c) 72(a,b,c) + 84(a,b,c) + 128(a,b,c) - 105(a,b,c) + 45(a,b,c)
= (a,b,c)(45+48-72+84+128)

The whole first 2 lines
" = (a,4a,7a) + (a,4a,8b) + (a,4a,9c) +
+ (a,5b,7a) + (a,5b,8b) + (a,5b,7a) + "
were removed, because these vectors were lin. dependent.

So, only vectors as (\alpha\vec{a}, \beta\vec{b}, \gamma\vec{c}) count as lin. independent. But, if you add some of the last "simplified" ones ...+(2b,4a,9c) + (2b,6c,7a) + (3c,4a,8b) + (3c,5b,7a) and add them, you don't get the original one?!

And, how is (a,6c,8b) same as 48(a,b,c)? Or where did 48 come from?



Addendum: How do you make strikethrough block of text? I've tried with HTML's <s> and <strike> and <del>, also found nothing with LaTeX.

 

[HallsofIvy]

What do you mean by "simplified"? What I would do is write it as a "linear combination" : (a+2b+3c, 4a+5b+6c, 7a+8b+9c)= a(1, 4, 7)+ b(2, 5, 8)+ c(3, 6, 9).

 

[courteous]

Instructions say (word-for-word):

Let \vec{a},\vec{b},\vec{c} be linearly independent. Simplify: (a+2b+3c, 4a+5b+6c, 7a+8b+9c).

PS: Where's the button for tex tags?