Thread: Multiple vectors, solving for each of them

This is a question from my calculus 3 class.

"Find u and v if u + 2v = 3i - k and 3u - v = i + j + k."

Alright, so I have me a pair of answers but they appear to be different than what the back of the book is telling me they are. I was also given a hint from the teacher that this is linear algebra and that I need to somehow eliminate variables. So I've sort of followed that advise.

The book says:
u = (5/7)i + (2/7)j + (1/7)k
v = (8/7)i + (1/7)j + (4/7)k

My work however shows similar answers but the constants are wrong. Here is what I did:

To get v:
I multiply ( u + 2v = 3i - k ) by -3.
Now I get ( -3u - 6v = -9i + 0j + 3k )
I add the other equation 3u - v = i + j + k and get ( -5v = 10i + j + 4k )
This ends up being v = -2i - (1/2)j - (4/5)k

To get u:
I multiply ( 3u - v = i + j + k ) by -2.
Now I get ( -6u - 2v = -2i - 2j - 2k )
I add the other equation u + 2v = 3i - k and get ( -5u = i - 2j - 3k )
This ends up being u = -(1/5)i + (2/5)j + (3/5)k

So as you can see, I get different answers. Where did I go wrong and did I even start this problem correctly to begin with?

Originally Posted by nautica17

This is a question from my calculus 3 class.

"Find u and v if u + 2v = 3i - k and 3u - v = i + j + k."

The book says:
u = (5/7)i + (2/7)j + (1/7)k
v = (8/7)i + (1/7)j + (4/7)k

u + 2v = 3i - k
6u - 2v = 2i + 2j + 2k
----------------------

7u = 5i + 2j + k

u = (5/7)i + (2/7)j + (1/7)k

looks like the book is correct (for u at least) ... recheck your work.

Looks like I put a plus instead of a minus on the 6u - v. :| I got the u correct now. Going to do v now.

Okay, got it working now.