it.hebe Posted February 15, 2009 Report Share Posted February 15, 2009 How would i solve this system of equations by hand of course625A -125B +25C -5D +E =4.5112A -34.3B +10.6C -3.25D+E=2.250.00391A -0.0156B +0.0625 -.25D +E=6.2592.4A + 29.8B +9.61C + 3.10 D + E= 1.40625A +125B +25C +5D +E =3.85Help would be very much appreciated Reply Link to post Share on other sites More sharing options...
Eternalx Posted February 15, 2009 Report Share Posted February 15, 2009 Try picking out two equations at do them at the same time then continue. That's if you're allowed. Reply Link to post Share on other sites More sharing options...
Hedron123 Posted February 16, 2009 Report Share Posted February 16, 2009 I cannot think of any other method other than elimination. It would be much more efficient to use the calculator as you don't get that kind of questions in the Exam. Reply Link to post Share on other sites More sharing options...
Chelleee Posted February 21, 2009 Report Share Posted February 21, 2009 i agree with hedron123, elimination is the only way. and can i ask what is this for?+ good luck with it, it looks like alot of work. Reply Link to post Share on other sites More sharing options...
Eternalx Posted February 21, 2009 Report Share Posted February 21, 2009 Oh also, it would be good if you can post the answer to this (the correct one) so i can check if i'm right or not ;D. It makes a good practice. Reply Link to post Share on other sites More sharing options...
Toffu-san Posted February 21, 2009 Report Share Posted February 21, 2009 How would i solve this system of equations by hand of course625A -125B +25C -5D +E =4.5112A -34.3B +10.6C -3.25D+E=2.250.00391A -0.0156B +0.0625C -.25D +E=6.2592.4A + 29.8B +9.61C + 3.10 D + E= 1.40625A +125B +25C +5D +E =3.85Help would be very much appreciatedI suppose that in the third line you have missed a "C".I've tried to do it and.... pfff It was infernal.Luck with it! Reply Link to post Share on other sites More sharing options...
the flying string Posted February 27, 2009 Report Share Posted February 27, 2009 It will be so fast if you could type this as a matrix on calculator Reply Link to post Share on other sites More sharing options...
Anonymouser Posted February 27, 2009 Report Share Posted February 27, 2009 (edited) Try doing it in row operations... an augmented matrix.Play with them till you get a triangle of zeros. It'll be hard work for a while but you'll eventually get it(for example if you subtract the last from the first, you'd cancel out three variables and be left with just another two.)(also, I suggest you multiply all the equations by 100,000 so you can get rid of any decimals or by 10,000 so you'd have just one decimal place left not 5 such as 0.00391)I do hope you're Math SL/HL cause if you're studies I don't know if you'd have taken augmented matrices.(btw you don't have to make it in triangle form, you just need to have one line with all variables, another with 1 less variable, and another with 2 less, and so on. It would still be solvable.)Example: Edited February 27, 2009 by Anonymouser Reply Link to post Share on other sites More sharing options...
Eternalx Posted February 27, 2009 Report Share Posted February 27, 2009 i just tried this, and man it was horrifying T___T i got nowhere, well i managed to get 2 rows with 0s and 1s Reply Link to post Share on other sites More sharing options...
moneyfaery Posted February 27, 2009 Report Share Posted February 27, 2009 i just tried this, and man it was horrifying T___T i got nowhere, well i managed to get 2 rows with 0s and 1sSolve for the variables in those rows and you can figure out the rest. I don't really see a point to this whole exercise though... the exam will never ask you something that stupid i.e. solving by hand. Just use your GDC. Reply Link to post Share on other sites More sharing options...
JohnPang.Tutor Posted February 28, 2009 Report Share Posted February 28, 2009 (edited) If you have studied the chapter "Matrix and its applications" , you may solve your system of equations by using the inverse matrix method.John Pang Edited February 28, 2009 by Aboo No advertising! Reply Link to post Share on other sites More sharing options...
Anonymouser Posted February 28, 2009 Report Share Posted February 28, 2009 (edited) Finding the inverse of a 5x5 matrix is literally impossible.Eternal, you don't need to have them 0's and 1's, you just need to have it in the form I described, no need to have the 1's. You can have 0.0453298239 as a coefficient for that matter and it'll still be solvable if you have the correct numbers of zeros.I have solved them for you. The results I got are pretty weird tho.Answers:The only way this can be solved by hand is using an augmented matrix and that would take some time. Edited March 1, 2009 by Aboo Reply Link to post Share on other sites More sharing options...
akimatsu123 Posted March 1, 2009 Report Share Posted March 1, 2009 Finding the inverse of a 5x5 matrix is literally impossible.Eternal, you don't need to have them 0's and 1's, you just need to have it in the form I described, no need to have the 1's. You can have 0.0453298239 as a coefficient for that matter and it'll still be solvable if you have the correct numbers of zeros.I have solved them for you. The results I got are pretty weird tho.Answers:The only way this can be solved by hand is using an augmented matrix and that would take some time.Why not use a calculator? Can't TI 84+ do inverse of 5x5 matrix? I can hardly imagine a 5 equation simultaneous equation with decimals showing up on paper 1. It's a waste of time to try to do this by hand, since you would never have to do it on the exam or in real life... Reply Link to post Share on other sites More sharing options...
Anonymouser Posted March 1, 2009 Report Share Posted March 1, 2009 The whole point of it (according to him) is to do it by hand... there's a billion ways to do it on computer/calculator. I did the one I've posted through Microsoft Math.it.hebe why do you want to solve it by hand? Reply Link to post Share on other sites More sharing options...
Eternalx Posted March 5, 2009 Report Share Posted March 5, 2009 Could someone teach me how to do this on the Ti-84 Calculator i tried and failed. Reply Link to post Share on other sites More sharing options...
Ongfufu Posted March 6, 2009 Report Share Posted March 6, 2009 (edited) Could someone teach me how to do this on the Ti-84 Calculator i tried and failed.625A -125B +25C -5D +E =4.5112A -34.3B +10.6C -3.25D+E=2.250.00391A -0.0156B +0.0625C -.25D +E=6.2592.4A + 29.8B +9.61C + 3.10 D + E= 1.40625A +125B +25C +5D +E =3.852nd MATRIX -> EDIT -> MATRIXPost by: Aboo 5x6 -> Enter the coefficients like this: 625 -125 25 -5 1 4.5 112 -34.3 10.6 -3.25 1 2.250.00391 -0.00156 0.0625 -0.25 1 6.25 92.4 29.8 9.61 3.10 1 1.40 625 125 25 5 1 3.85After this painful step, press 2nd QUIT -> 2nd MATRIX -> MATH -> go down to B: rref( -> ENTER -> 2nd MATRIX -> NAMES 1: Post by: Aboo -> ENTER -> ENTERThis should solve it:1 0 0 0 0 .02379174390 1 0 0 0 .00677548550 0 1 0 0 -.67713927590 0 0 1 0 -.23438713760 0 0 0 1 6.233641964Therefore, A=.0237917439, B=.0067754855, C=-.6771392759, D=-.2343871376, E=6.233641964. Edited March 8, 2009 by ongfufu Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.