## Dido525 2 years ago For which values of a will the following system have no solutions? Exactly one solution? Infinitely many solutions? x+6y+4z=2 4x+25y+31z=7 2x+13y+(7+a^2)z=-1+a I tried elimination but that didn't work... I don't know how I would use matricies for this either.

1. Dido525

@zepdrix @UnkleRhaukus @saifoo.khan

2. BluFoot

First, turn it into a matrix!

3. Dido525

Umm okay...

4. Dido525

Okay.

5. BluFoot

Here do you understand how I got this? [1 6 4 2 ] [4 25 31 7 ] [2 13 7a^2 -1+a]

6. Dido525

Yeah I do.,

7. BluFoot

Actually it should be 7+ a^2 so [1 6 4 2 ] [4 25 31 7 ] [2 13 7+a^2 -1+a] OK now do you know how to row reduce this?

8. Dido525

Yeah but what about that a?

9. Dido525

@BluFoot

10. BluFoot

We're going to work with that later. It shouldn't get in the way because it's in the bottom row.

11. Dido525

I have to still manipulate it like normal correct?

12. BluFoot

Yes. So first try to get 0s in the first column

13. Dido525

|dw:1358201674350:dw|

14. Dido525

No wait. That's not right.

15. BluFoot

Nope :P OK I'll give you the first step. Subtract 4 times the first row from the second row.

16. Dido525

Sorry about that. This Linear algebra is fun but totally new to me.

17. BluFoot

Nono don't be sorry, it's not an easy course at all! I just took it last semester so it's still fresh in my mind. Everybody said it would be easier than calculus, but I think they were crazy, linear is way harder...

18. Dido525

If we went 4* Row 1 we get -3 in the 2nd row for the first term.

19. BluFoot

OK so subtracting 4*R1 from R2 you get. 4*R1 = 4 * [ 1 6 4 2 ] = [4 24 16 8] R2 - 4*R1 = [ 4 25 31 7 ] - [4 24 16 8] = [0 1 15 -1] Make sense? So now your matrix looks like [1 6 4 2 ] [0 1 15 -1 ] [2 13 7+a^2 -1+a]

20. Dido525

Wait... Oops.

21. Dido525

KK. Let me try myslef again.

22. Dido525

I was looking at another Matrix >.< .

23. Dido525

Kk. I got what you got. Now I would go and divide Row 3 by 2 right?

24. Dido525

@BluFoot

25. BluFoot

Nope you would do Row 3 - 2 times Row 1. You want to get a 0 in the first spot of row 3

26. Dido525

Okay, that works too.

27. Dido525

For row 3 I got [0 1 5+a^2 -3+a]

28. Dido525

Wait, third colums in wrong.

29. Dido525

[0 1 -1+a^2 -3+a]

30. BluFoot

Perfect very good! :D [1 6 4 2 ] [0 1 15 -1 ] [0 1 -1+a^2 -3+a] almost there. Now you want to get a 0 in row 3 column 2

31. Dido525

[0 1 -1+a^2 -5+a] . Woudn't it be that actually?

32. BluFoot

Oops yes! Missed that, just skimmed over it. Good job noticing that. So now it's [1 6 4 2 ] [0 1 15 -1 ] [0 1 -1+a^2 -5+a]

33. Dido525

K I got [0 0 -16+a^2 -4+a]

34. BluFoot

Perfect! OK I have to go so I'll just go over the answer and stop giving you hints. If you want to guess for yourself do that first then read my answer! [1 6 4 2 ] [0 1 15 -1 ] [0 0 a^2-16 a-4] is equal to (by difference of squares) [1 6 4 2 ] [0 1 15 -1 ] [0 0 (a+4)(a-4) a-4] So you've got (a+4)(a-4) * z = a-4 If the left equals 0, but the right equals a number, then there is no solution because: 0*z=1 has no solution. So when does the left equal 0 but the right equal a number? When a = -4. Then 0*z = -8, no solution. If they both equal a number, let's say a=5 then [1 6 4 2 ] [0 1 15 -1 ] [0 0 9 1 ] You now have 3 pivot columns, which means that there is only 1 solution. If they both equal 0, then you will have a free variable in the third column and you will have infinite solutions: [1 6 4 2 ] [0 1 15 -1 ] [0 0 0 0 ] When do they both equal 0? When a =4, infinite solutions. Good luck in linear algebra! It's a fun class, but make sure you don't fall behind, it gets really hard!!

35. BluFoot

36. Dido525

WHat's a Pivot column?

37. Dido525

And thanks for the encouragement :) .