The vetctor space is given by:
r = {1,1,2,3}t1 + {1,4,2,1}t2
So:
x = t1 + t2
y = t1 + 4t2
z = 2t1 + 2t2
w = 3t1 + t2
Letcs take the first and fourth equations.
x = t1 + t2
w = 3t1 + t2
From here we get t1 = (w - x)/2, t2 = (3x - w)/2.
Now, the second and third equations.
y = t1 + 4t2
z = 2t1 + 2t2
From here we get t1 = (2z - y)/3, t2 = (2y - z)/6.
So, combining all that, we get two equations for x, y, z, w:
(w - x)/2 = (2z - y)/3
(3x - w)/2 = (2y - z)/6
Simplifying a bit, we get:
3x - 2y + 4z - 3w = 0
9x - 2y + z - 3w = 0
#How do I solve this?
stable laurel
Of course, this answer isn't unique, but all answers will be systems of 2 equations in 4 variables.
slow sedge
This is the solution I had
slow sedge
Thank you
stable laurel
Why did you delete it?
slow sedge
Because I'm scared my school is watching
stable laurel
Huh...
Well, alright...
I just wanted to tell you that you can simplify your equations by gettid rid of rational coefficients.
So, multiply the first one by -1 and the second by 3.
slow sedge
The first one by -1?
stable laurel
Well, yeah. I just prefer when the leading coefficients are positive.
Oh, and then the second one by -3.
stable laurel
That isn't required, of course. I just find it aesthetically pleasing.
stable laurel
Wouldn't that be changing the equation though?
Well, are x + y = 0 and 2x + 2y = 0 really that different?
slow sedge
Ohhh
stable laurel
You can multiply equations by any nonzero constants, it doesn't change them.
slow sedge
Man thank you
stable laurel
You're welcome!
slow sedge
Do you mind telling mehow you isolated for t1 and t2 in your solution? (t1 = (w - x)/2, t2 = (3x - w)/2)
stable laurel
Do you mind telling mehow you isolated for t1 and t2 in your solution? (t1 = (w ...
Oh, I just solved the system in each case.