## anonymous one year ago help me plz again

1. anonymous

2. anonymous

@Nnesha

3. anonymous

@paki

4. anonymous

@dan815

5. anonymous

@jim_thompson5910

6. jim_thompson5910

how far did you get? were you able to set up the matrix?

7. anonymous

i was not i was so confused on this like i dont know where to start

8. jim_thompson5910

well you first pull out the coefficients of each equation. Do you see what they are?

9. anonymous

i do not sorry

10. jim_thompson5910

they are simply the numbers in front of the variables

11. jim_thompson5910

example: 4x has the coefficient of 4

12. jim_thompson5910

the -y is really -1y

13. anonymous

so four and five right

14. jim_thompson5910

along the first equation, what are the 3 coefficients?

15. anonymous

4 5 and 1x right

16. jim_thompson5910

17. jim_thompson5910

not column

18. jim_thompson5910

do you see what I mean?

19. anonymous

ohh ok i got you hold on

20. anonymous

so 4 and three for the first one right

21. jim_thompson5910

22. jim_thompson5910

the first equation is really 4x - 1y + 3z = 12

23. anonymous

ohh yeah sorry lol

24. anonymous

so the seciond one is 1x+4y+6z=12

25. jim_thompson5910

-32 on the right side, not 12

26. anonymous

opps my bad i looked at it wrong haha i acciedently looked up at the end sorry

27. anonymous

5x+3y+9z=20 for the last

28. jim_thompson5910

So if we just focus on the coefficients on the left side, we have this 3x3 matrix $\left[\begin{array}{ccc} 4 & -1 & 3\\ 1 & 4 & 6\\ 5 & 3 & 9 \end{array} \right]$

29. jim_thompson5910

agreed? or no?

30. anonymous

yes i agree

31. jim_thompson5910

the next matrix is simply the matrix of variables written as a column like this $\left[\begin{array}{ccc} x\\ y\\ z \end{array} \right]$

32. jim_thompson5910

and then finally we have a third matrix of the values on the right side of the equation the next matrix is simply the matrix of variables written as a column like this $\left[\begin{array}{ccc} 12\\ -32\\ 20 \end{array} \right]$

33. jim_thompson5910

putting this all together, we form this matrix equation $\left[\begin{array}{ccc} 4 & -1 & 3\\ 1 & 4 & 6\\ 5 & 3 & 9 \end{array} \right] * \left[\begin{array}{ccc} x\\ y\\ z \end{array} \right] = \left[\begin{array}{ccc} 12\\ -32\\ 20 \end{array} \right]$ making sense so far?

34. anonymous

yeah

35. jim_thompson5910

next we form an augmented matrix by ignoring the variables temporarily. We just combine the first and third matrix like this $\left[\begin{array}{ccc|c} 4 & -1 & 3 & 12\\ 1 & 4 & 6 & -32\\ 5 & 3 & 9 & 20 \end{array} \right]$ the vertical bar separates the two matrices

36. anonymous

makes since now

37. jim_thompson5910

this is where gauss jordan elimination begins

38. anonymous

ok

39. jim_thompson5910

do you know how to do gauss jordan elimination? or no?

40. anonymous

i really dont i have treid and i keep faling at it this is whay im here now hahaha

41. jim_thompson5910

that's ok

42. jim_thompson5910

The first thing I would do is swap rows 1 and 2. We denote this with notation R1 <--> R2. I'm doing this to make the element in row1,column1 be a 1 $\left[\begin{array}{ccc|c} 4 & -1 & 3 & 12\\ 1 & 4 & 6 & -32\\ 5 & 3 & 9 & 20 \end{array} \right]$ $\left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 4 & -1 & 3 & 12\\ 5 & 3 & 9 & 20 \end{array} \right] \begin{array}{c} \phantom\\ R_1 \leftrightarrow R_2\\ \phantom\\ \end{array}$ I couldn't get the thing to line up properly, but the R1 <--> R2 should either be in row 1 or row 2

43. jim_thompson5910

let me know when you're ready for the next step

44. anonymous

OK GIVE ME A SECOND TO GET THIS DOWN sorry for the caps i for got i had them on

45. anonymous

46. jim_thompson5910

Now we want to make that 4 in row2,column1 to be a 0. We do this by computing R2-4*R1 and replacing all of R2 with that result $\left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 5 & 3 & 9 & 20 \end{array} \right] \begin{array}{c} \\ R_2 - 4R_1 \rightarrow R_2\\ \\ \end{array}$

47. anonymous

OK GOTCHA

48. jim_thompson5910

Next we want to make that 5 in row3,column1 to be a 0. We do this by computing R3-5*R1 and replacing all of R3 with that result $\left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & -17 & -21 & 180\\ \end{array} \right] \begin{array}{c} \\ \\ R_3 - 5R_1 \rightarrow R_3\\ \end{array}$

49. anonymous

ok i got that

50. jim_thompson5910

We could divide the second row by -17 to make that first '-17' in that row equal to '1', but I'm going to skip that step. Instead I'm going to subtract R2 and R3 and replace R3 R3-R2 --> R3 $\left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & 0 & 0 & 40\\ \end{array} \right] \begin{array}{c} \\ \\ R_3 - R_2 \rightarrow R_3\\ \end{array}$

51. jim_thompson5910

The last row of this matrix $\left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & 0 & 0 & 40\\ \end{array} \right]$ tells us that 0x + 0y + 0z = 40 which essentially turns into 0+0+0 = 40 and that turns into 0 = 40

52. anonymous

wow that was alot

53. jim_thompson5910

but 0 = 40 is NEVER true regardless of what x,y,z are. Since we have a contradiction, we have an inconsistent system that has NO solutions at all.