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Linear Algebra, simplifying large matrices

  • Thread starter maximade
  • Start date
  • #1
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Homework Statement


Find all solutions using Gauss-Jordan elimination:
[ 0 0 0 1 2 -1 l 2
1 2 0 0 1 -1 l 0
1 2 2 0 -1 1 l 2]

Homework Equations


Switching rows,
able to scale any row
able to add non zero multiple to row

The Attempt at a Solution


What I did was switch the 1st and 3rd row, then i tried my best to get it as close to rref form.
I ended up getting:
[ 1 2 0 0 1 -1 l 0
0 0 -2 0 2 -2 l -2
0 0 0 1 2 -1 l 2]

I honestly can't think of a way to reduce it anymore, not to mention that I have no idea how to even put this in "solution" form.
 

Answers and Replies

  • #2
33,507
5,192

Homework Statement


Find all solutions using Gauss-Jordan elimination:
[ 0 0 0 1 2 -1 l 2
1 2 0 0 1 -1 l 0
1 2 2 0 -1 1 l 2]

Homework Equations


Switching rows,
able to scale any row
able to add non zero multiple to row

The Attempt at a Solution


What I did was switch the 1st and 3rd row, then i tried my best to get it as close to rref form.
I ended up getting:
[ 1 2 0 0 1 -1 l 0
0 0 -2 0 2 -2 l -2
0 0 0 1 2 -1 l 2]

I honestly can't think of a way to reduce it anymore, not to mention that I have no idea how to even put this in "solution" form.
So far, so good, but you can go one step further by replacing the 2nd row by (-1/2) of itself.

That gives you
[1 2 0 0 1 -1 l 0]
[0 0 1 0 -1 1 l 1]
[0 0 0 1 2 -1 l 2]

Now it's in RREF form, with each leading entry being 1.

What you have is a matrix that represents three equations in six variables. A system like this typically has an infinite number of solutions.

To write the solution, solve for each variable that has a leading entry: x1, x3, and x4. The other three variables are free variables that can have any value.
 
  • #3
27
0
Cool, I didn't know that the rref form doesn't need to have straight diagonal "1"s.

As for the solution, would it simply be:
a+2b+e-f=0
c-e+f=1
d+2e-f=2?
 
  • #4
33,507
5,192
I would go further and solve for the first variable in each of your equations.

a = -2b - e + f
c = e - f + 1
d = -2e + f + 2
b, e, f are arbitrary
 

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