I applied Gauss Jordan on a matric and result what i get is
x+2y=3
y=0
z=free
How to find three intersection points for this system? Kindly help
matrix sorry
Well, clearly there is no unique solution involving z.
Apparently x=3 and y=0
So, any point of the form (3,0,z) will satisfy the equations.
Thanks
To find the three intersection points for the given system of equations, we need to solve it using the given results from applying the Gauss-Jordan elimination.
The given system of equations can be rewritten as:
1) x + 2y = 3
2) y = 0
3) z = free
From equation 2), y = 0, we can substitute this value of y into equation 1) to solve for x:
x + 2(0) = 3
x = 3
So, we have the first coordinate of the intersection point as (x, y, z) = (3, 0, z).
Since z is free, we can assign different values to it and find the corresponding values of x and y.
For example, if we choose z = 0, the intersection point becomes (3, 0, 0).
Similarly, if we choose z = 1, the intersection point becomes (3, 0, 1).
You can choose any value for z to get different intersection points on the x-y-z coordinate plane.