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find all solutions x1, x2, x3 of the equation vector b= (x1 times vector v1) + (x2 times vector 2) + (x3 times vector 3) where

vector b=
{-8}
{-1}
{2}
{15}

vector 1=
{1}
{4}
{7}
{5}

vector 2=
{2}
{5}
{8}
{3}

and vector 3=
{4}
{6}
{9}
{1}

I know these vectors can be simplied to the following equations, but i do not know how to solve 3 variables with 4 equations:

x1 + 2 x2 + 4 x3 = -8
4 x1 + 5 x2 + 6 x3 = -1
7 x1 + 8 x2 + 9 x3 = 2
5 x1 + 3 x2 + 3 x3 = 15

ANY HELP WOULD BE GREATLY APPRECIATED

2007-09-18 11:48:37 · 1 answers · asked by MARK 2 in Science & Mathematics Mathematics

1 answers

Okey Dokey. The first thing you want to do is put this into Matrix Form

|1 2 4 -8 |
|4 5 6 -1 |
|7 8 9 2 |
|5 3 3 15 |

Now if you put this Matrix into Reduced Row Echelon Form (you can either work that out by hand or use MATLAB or Mathematica if you are allowed to), you'll end up with the identity matrix. This tells you that those four column vectors that make up the above matrix are linearly independent and that tells you that NO combination of the 3 vectors can equal b. This tells you there are no solutions.

That is how you can solve this problem using linear algebra.

If you didn't follow that, you can still show that is the case by using the first equation to put one variable in terms of the other 2, use the second equation to put get one variable in terms of one other one, then use the third to solve that system of equations. But then you'll still have to make sure that this checks out with the fourth equation (which it won't) so that is another way to show there is no solution to this system of equations. I'm not going to do that for you, but I think you should be ok from here. Good luck!

2007-09-18 11:53:45 · answer #1 · answered by Anonymous · 0 0

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