Manipulating Matrices: Elementary Row Operations and Gauss-Jordan Elimination
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- čas přidán 9. 10. 2018
- Now that we know how to represent systems of linear equations by using matrices, how can we solve those systems while in matrix notation? The easiest way is called Gauss-Jordan elimination, so let's learn how to do it!
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Dude, rly? You're posting videos at the same pace as I'm learning these things in college. That's awesome.
Same. Same.
Professor Dave, your videos are well done. I would make one suggestion, when you give the final problem on your matrices to practice, don't just show the answer, but show how you solved it. It is important to learn and reinforce the process. Thank you!
1 1 1 5 1 1 1 5
2 3 5 8 => we multiply the 1st row by 2 so we can substract the first from second row and we get 0 1 3 -2
4 0 5 2 4 0 5 2
=> than we (note that we don't change the first row) multiply the first row by 4 and substract the 1st row from 3rd
and we get:
1 1 1 5
0 1 3 -2
0 -4 1 -18
=> now we can multiply the second row by 4 and combine the 3rd and second row so we get
1 1 1 5
0 1 3 -2
0 0 13 -26
here we can devide the whole 3rd row by 13 and we get 0 0 1 -2 (z = -2) and knowing that we can get x and y
Its essential to work with 0 and 1 so you make your work easier
You have a gift, honestly most professors can’t help but ramble on and talk on circles making this more confusing than they should be. You say only what is need in a clear and concise way. You’re videos are my absolute favorite.
anyone here during the 2020 pandemic teaching themselves college at home
Not really on our own.... love from Somalia🇸🇴
Yes ma'am.
yes...
teaching myself linear before i start linear in 3 weeks
hell yeah 😂
For anyone else who had a hard time with this, 2 things really helped me to start understanding the strategy for elimination:
1. Dave says so in the video, but it went right over my head: Don't worry about getting 1 values in the matrix, focus only on getting 0 values. Once all but the rightmost value and one other value in a row are zeroed, you can divide both values by the coefficient non-zero value, which will convert the row to RREF and provide a solution.
2. It helps me to start zeroing from the top-right or bottom-left corner of the coefficient matrix. It almost feels like a game or a puzzle: Getting those corner values zeroed provides a solid base to start zeroing horizontally or vertically without accidentally limiting yourself to operations that would change a previously created zero.
Anyhow, hope that helps. If anyone has any other strategies, I'd love to hear them. Thanks for the video, Dave! Looking forward to more!
@johnryan7661
Honestly, these are some nice reminders for algebra in general... Good on you! 👏🏾
Man, studying these (relatively) advanced Math courses in College really reveals which fundamentals in Algebra you didn't understand since every advanced concept is just built on the basic foundations. In my case, I didn't really know much about the elimination method (I always solve using substitution method) so I had to learn elimination method before rewatching this video.
YO! the example you did much easier than the question at the end...
professor dave this is hard i hate teaching myself math but ur videos help thanks for ur coolness man
This clears a lot of things up, thanks.
matrices,learning now! thanks professor dave! 😊 please upload videos on determinants also🤔🤔
Excellent explanation, thank you!
Sir, you are a good teacher who knows how to send message to the recipient. Great.
Thank you Professor!
Thanks for this, I needed this because I have an assignment due tomorrow.
Your the best professor.
this problem took me more than two hours. but it felt great to figure it out.
10k views and 0 dislikes! that's insane dude. he's also better than my +150$k paid professor.
i hope you know you caused the one dislike he did get now lol. not by me
Now 5 dislikes come from jealous professors. 😁
Thanks!
Thanks for the videos. :D
I was wondering how many solutions there are in your example?
Is it infinitely many solutions because the rank(3) is less than columns(4)?
You need the columns in the coefficient matrix, not the augmented matrix as mentioned in the video at 8:40.
I did not know why we augment the rhs column. Now I do. Thank you.
Wow what a life saver.
sir I would have failed my class is it were not for you. Thank you
Bro, you are a genius, thank you very much!!!
I’m interested and scared as to where this series will depart from the matrices stuff learned in algebra 2 and pre Calc to the scary linear algebra stuff
Sir are there any mathematical objects that can't be added to themselves?
Sir how are scalars represented graphically?
Solution:
R3=R3-4R1
R2=R2-2R1
R3=R3+4R2
R3/13= Z=-2
R1=R1-R2
R1=R1+2R3 X=3
R2=R2-3R3
Y=4
Thank you 😁🙏❤️🤝
thank you so much
thank you !!
hello I'm lost after the R3/13, it gives me weird results that block me from going on? :/
thank you I was completely stuck on that one!
Thanks dave
Tne solution must be
X=39/5
Y=-16/5
Z=2/5
thank you
you are a saviour
Thanks
3:00 is it up to us to decide which equation to keep and which one to change to the resulting one?
yes
God bless you Dave, we love you ❤️
i understood everything, but the X Y Z arond minute 6:30, what makes the second row Z and the third row Y? Isn't it supposed to be organized like X first row and the second should be Y instead of Z?
Switching rows doesn't change the system of equations, but makes elimination easier. In this case, the 2nd row is z=0 and the 3rd row is y+z=5.
iam wondering the same thing why is it allowed to be happened???
If youre feeling stuck on the comprehension, remember that only the bottom left and top right need to become 0 (ignore the last column with the answers) and that you only need to operate on one row at a time. You can do it!!
Sir how are scalars represented graphically? plzz give a reply
scalars are just numbers so we wouldn't really graph them
This is what it was? Seriously? Been staring at my textbook for an hour to find out its the same exact stuff we did in high school 🙃
can some one please upload how to solve the last equation of the video its harder than the others
i also cant find the solutions :(
Can someone explain how the challenge at the end is solved? I thought I understood this very well but found it to feel like a rubix cube where all my "moves" are unproductive, as even though I can get a zero in a spot, I end up losing it when I get a zero in a different spot.
Now that you've mentioned it i realized it's akin to solving a rubik's cube lol
Anyways this is how i done it:
R2=R2-2(R1)
R3=R3-2(R2)
R1=R1-R2
R3=R3+6(R2)
R3=(R3)/13
R1=R1+2(R3)
R2=R2-3(R3)
Tips: Make [1 0 0] first in the first column, and then try to make [0 1 0] in the second column, and then [0 0 1] in the third column. Very similar to solving a rubik's cube where we start by solving the bottom first and moving above, in here we start by the left side first then moving to the right side
Alloh rozi bosin, borakanu norm darslayam
Im super confused on how to pick which equaton to perform first on the comprehension part
Can we reduce the equations once we get just 2 vars in the equations and find values by getting relation between the vars like x=2z etc or we need to do only matrix operations until we arrive at 1,0,0 / 0,1,0 etc format?
Depends on who's asking you to do it. It is a perfectly valid solution to the original problem, to do what you propose, and if it were up to me, I'd give full credit for your solution method. I.e. stopping part way with Gauss-Jordan elimination (GJE) once you determine one of your variables, and simply using relationships among the variables to find the others. So as long as the person asking doesn't care about the method, it is perfectly valid to do what you propose.
By contrast, if you are in a class where you are learning GJE, they are probably expecting you to take the row operations all the way, and solve for all of the variables with the method you are being tested upon. If anything, I don't really see the advantage in doing what you propose, because it is ultimately the same work to finish the GJE completely, and what you propose would require another mental exercise to do.
why cant i understand all the subtracting stuff
Is there anyone with the solution steps to the last matrix [1 1 1 5, 2 3 5 8, 4 0 5 2] on this video?
R3=R3-4R1
R2=R2-2R1
R3=R3+4R2
R3/13= Z=-2
R1=R1-R2
R1=R1+2R3 X=3
R2=R2-3R3
Y=4
Why is it that I found x=123/13 , y= -74/13 and z= 16/13????? Somebody please help me out. I have an exam in two daysss
You replace the row being subtracted
That should be 0 1 2 5 not 0 1 1 5 at 5:43 because 11 - 3 * 3 = 2
It's 11 - 7 - 3 = 1.
soln:
R2- 2R1
R1 - R2
R3 - 4R1
R3 / 13 => z = -2
R1 + 2R3 => x = 3
R2 - 3R3 => y = 4
i know that musicians understand math well.
Professor did the whole maths topics has covered?.... Nah where is determinants pls sir make a vedio on that tooo
buddy relax, more math is coming
@@ProfessorDaveExplains Sir is it will be with in 2months?
Recommending this to all my friends
The Legends are here 1 day before exam 😅
That’s me
its not really clear WHY adding two rows produces the same linear equations.
For the same reason you can add or subtract two equations in general. It's simply combining two mathematical facts of the given information, into one mathematical fact.
the practice problem at the end is disgusting
It's easy the first row is a big clue already
i hate matrices
You will get accustomed to them. Matrices save you from writing out x, y, z, etc.
At 8:15 ur property is incomplete ..all the entries **below** the leading one in the column must be 0
Row echelon form and reduced row echelon form are different. The video is correct as is.
@@ProfessorDaveExplains I see ...what exactly are the differences between the two?
Simply ones on the diagonal vs. ones and zeroes for the rest of the column so each variable is isolated.
@@ProfessorDaveExplains didn't quite understand
Bennettians 💋💋
Thanks!
Really appreciate how you make content vastly more digestible than others. Amazing work!