I've started university lately and i'm currently taking a math class in linear algebra. I'm going well so far but i'm having some issues with solving a system of linear equations when there is a unknown value in the system of equation. I must be missing a point somewhere. but here i go.
x - 2y + 3z = 1
-3x + y - 3z = 2
-2x - y + (kČ -2k)z = k +3
I have to find k in order for the system to have at least one solution.
Any help would be appreciated as the more i try to solve it the more i get confused.
I think i'm getting confused by the (kČ-2k) section of the equation.
solve this one
if determinant of that is than sys have unique solution ....
1° when and if at least one of than solutions are unique and given
2° if and if at least one of than system of equations have no solutions
3° if and if than there can be :
a) all sub determinants of D are zeros, and coefficients with the unknowns are proportional there are 2 cases :
- all independent members of system are proportional with coefficient with unknowns than there are infinity of solutions
- all independent members of system are not proportional with coefficient with unknowns than there are no solutions
b) if at least one sub determinant of D are different from zero than there are infinity of solution
Edit: ah sorry ... I assumed that you are familiar with this, because you post question on university sub forum .... sorry
Maybe my post should be in pre-u section of this forum. I believe that the american and canadian school system is different. University for us i believe is college in the states though i'm not sure. Here we have the equivalent of elementary->highschool->college->University.
Express x from the first equation and insert it to 2 and 3.
You have two equations with y and z.
Express y from one equation and insert it to the third.
Finally you have
a(k)z + b(k) =c(k).
Now examine it.
1) if a(k)<>0 there is only one solution.
2) if a(k)=0 and b(k)=c(k) there are many solutions.
3) if a(k)=0 and b(k)<>c(k) there is no solution.