## angela210793 Group Title Find all the solutions possible of the system ...using matrices 2 years ago 2 years ago

1. wasiqss Group Title

system is hidden LOl

2. Kreshnik Group Title

wait a second.. she must be writing it !!

3. angela210793 Group Title

|dw:1334243246880:dw|

4. wasiqss Group Title

anxhela , you want it to be done by cramer's rule or gauss elimination

5. angela210793 Group Title

i did |dw:1334243367057:dw| i know wassiq...but u can't use cramer when determinative equals 0

6. angela210793 Group Title

|dw:1334243466671:dw|

7. wasiqss Group Title

ahhh yes, determinant is zero . we must go for gauss elimination

8. wasiqss Group Title

do you know gauss, i can do that

9. angela210793 Group Title

i did |dw:1334243615536:dw| krocker-capelli but i got

10. angela210793 Group Title

i know Gauss but we have a k in there

11. wasiqss Group Title

good :), i havnt done krocker one til yet, but yes k would have been a headache in gauss

12. Zarkon Group Title

k has to be 1

13. Zarkon Group Title

if you want a solution

14. angela210793 Group Title

how did u find it

15. TuringTest Group Title

Zarkon is the only one I won't chastise for just giving answer since they are so provocative ;)

16. Zarkon Group Title

-2 times row one = row 3

17. wasiqss Group Title

Zarkon if k=1 , then game is over :D

18. Zarkon Group Title

no...if $$k\ne 1$$ then there is no soluition.

19. Zarkon Group Title

solution

20. angela210793 Group Title

u see..i tried Gauss..and i got at the row3 0 0 0 2 where did i got wrong O.o

21. Zarkon Group Title

$\left.\begin{matrix}x-y+z=k \\ -2x+2y-2z=-2\end{matrix}\right.$ multiply first by 2 $\Rightarrow$ $\left.\begin{matrix}2x-2y+2z=2k \\ -2x+2y-2z=-2\end{matrix}\right.$ add you get $$0=2k-2\Rightarrow k=1$$

22. wasiqss Group Title

NIce work zarkon

23. phi Group Title

If you put the augmented matrix in reduced row echelon form you get $\left[\begin{matrix}1 &0 & 1&1 +k/2 \\ 0 & 1 &0 & 1-k/2 \\ 0 & 0 & 0&2k-2 \\\end{matrix}\right]$

24. phi Group Title

as zarkon noted, k must be 1 for there to be a solution. However, note that there are an infinite # of sols if k=1, because you have a non-zero null space

25. angela210793 Group Title

the system would have infinite solutions for r(A)=r(Ab) and no solutions for r(A)<r(Ab) how do u find the range of Ab when u have k in there..r(A)=r(Ab) if k wasn't 1....sorry mozilla froze -_-

26. angela210793 Group Title

ok...thanks

27. phi Group Title

The complete solution is $\left(\begin{matrix}3/2 \\ 1/2 \\ 0\end{matrix}\right)+c\left(\begin{matrix}1 \\ 0 \\ -1\end{matrix}\right)$ where c is arbitrary