# Thread: Solving a system of linear equations.

1. ## Solving a system of linear equations.

Hi Everyone,

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 + (k² -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 (k²-2k) section of the equation.

thanks.

2. Originally Posted by Hyrules
Hi Everyone,

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 + (k² -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 (k²-2k) section of the equation.

thanks.
find determinant of the system ... and on depending of the value of "k" you will have one, none or infinity of solutions ...

3. ouch... Thanks for the answer unfortunatly the determinant is the next chapter in the math book. I have to do it some other way or should i say the long way.

4. Originally Posted by Hyrules
ouch... Thanks for the answer unfortunatly the determinant is the next chapter in the math book. I have to do it some other way or should i say the long way.

$\displaystyle D= \begin{vmatrix} 1&-2&3\\ -3&1&-3\\ -2&1&(k^2-2k) \end{vmatrix} = 1\cdot 1\cdot (k^2-2k) + (-2)\cdot (-3) \cdot (-2) + 3\cdot (-3) \cdot 1 - \big{[} 3\cdot 1\cdot (-2) + (-2)\cdot (-3)\cdot (k^2-2k) + 1 \cdot (-3) \cdot 1\big{]} = . . . .$ solve this one

if determinant of that is $\displaystyle D \neq 0$ than sys have unique solution ....

1° when $\displaystyle D\neq 0$ and if at least one of $\displaystyle D_x\vee D_y\vee D_z\vee \neq 0$ than solutions are unique and given

$\displaystyle \displaystyle x = \frac {D_x}{D} , y= \frac {D_y}{D} , z = \frac {D_z}{D}$

2° if $\displaystyle D= 0$ and if at least one of $\displaystyle D_x\vee D_y\vee D_z\vee \neq 0$ than system of equations have no solutions

3° if $\displaystyle D= 0$ and if $\displaystyle D_x= D_y= D_z= 0$ 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

5. 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.

6. 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.

7. great that was the answer i was looking for ! Thanks.