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Homogeneous Systems of Linear Equations - Trivial and Nontrivial Solutions, Part 2
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- čas přidán 7. 08. 2024
- Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) / patrickjmt !! Homogeneous Systems of Linear Equations - Trivial and Nontrivial Solutions, Part 2. In this video, I show how to find solutions to a homogeneous system of linear equations that has nontrivial solutions.
patrickJMT = youtube legend...
TheCireMC F**king right.. He is the holy water for the devil math..
He has helped many students pass their math classes
Agree
@kai23430 thanks, more linear algebra coming. thumps up if you like them! it helps me : )
I have been watching your videos lately, you are a great instructor as you can explain such topics easy and clear. You're amazing.
Thank so much. You make it very simple to understand by going through each step and not just assuming we know the steps. Very informative and a life saver. Keep up the good work and thanks.
I wish I could find you much more earlier to learn everything clearly.Seriously dude you are rock! Unfortunently my exam is tomorrow so I am gonna need luck very much :D Thanks for your clear expression :)
Absolutely! He said at the beginning its like grass grow. But he still does it and continues to which is why its awesome! Thank you for your huge patience!
thank you man...wouldn't have understood this for my test without your help. it's just right to appreciate you. God bless you man
my pleasure :) thanks for subscribing
PATRICK please post about non-homogeneous systems of linear equations also .....thanks
YOU'RE A LIFE SAVER omg thanks a lot man
Very well explained! Didn't have to read the book. Thanks a lot! :)
THIS GUY ALWAYS COMES CLUTCH !!!
amazing, i learned in just one session, just subscribed
. thanx
Thanks for the video! Helps a ton!!!
Thank you very much for that awesome video! It helped me a lot! Best greetings from Switzerland :-)
12 years and still very helpful!!🫡
Wow this video helped me so much. Thanks Patrick
Clear and easy to understand. Excellent!
Great Work! You'll be a Professor in no time.
@Gytax0 well, this example did
My handout made sense with your tutorial. Thanks 😊
Excellent explanation! Your videos have helped me from cal 1
very nice tutorial ! Thank you Patrick !
Thank you.
But just a question to see if I understood the "Linear dependance" part well.
These vectors are dependent .. right ?
This is a good review in linear algebra for me
good work and keep it up :)
THANK YOU, YOU'RE A LEGEND
you explained this better in 10 minutes then my prof in 45 minutes
@HaidenDaggett happy to help : )
Could you make a video on solving systems of linear equations with parameters in?
its still hard for me to control the nontrival solutions .
But this video help me a lot!
THX!
Used Khan Academy til I found this. Both were Godsend, but prefer this =) Helping me pass difficult first year Math! Thank you
i love the sound of a sharpie on a fresh piece of paper
l love your channel it is really a good explanation you are the best
Glad you think so!
Dude you are so much better than my professor!
THANK YOU SO MUCH!!! I'm not really a fan of my professor's lectures, so this is so helpful!!
thank you
that's so helpful. Much better than my uni 's professor who can barely speak english jesus!!!
Very helpful! Thankyou!
very easy and clear
Thank you sir!
Thank you sir 😊
For this video
what is the reason that we turned this system into RREF from but ones in the previous videos were not? They were upper triangular systems
ANybody explains for me?
Great work and thanks
Thank you so much
@Gytax0
Systems have either a) No solutions (parallel or skew lines/planes/hyperplanes) (I dont believe is possible in a homogeneous system, since they all pass through the origin), b) One solution (a single intersecting point between lines), c) Infinitely many solutions (either equivalent lines, or the intersection of planes or hyper-planes)
Mathematician Patrick SMART thank you
Saved my life
that is the third time i have read that today :)
as i told the other two people: in that case, you owe me one!
Patrick I love you. alot.
6 b 4^&__^???*(&5€_=
subbed and liked. thank you good sir
Thanks for the video! it helps a lot~~~~
Thanks for the help
Great work!! Thank you so muck :$
Thanks 😊
woooow good teacher i had lean a lot
good job! bless you!
great video
Thanks Good Explanation
Dear profesor ,Thanks to help me for this video
could I ask a question ?
I think nontrivial solution is: if a lineer algebric system's coefficant matrix determinant is equal zero
if the system's rigth sides are NOT zeros
can we say it has nontrivial ?
thanks for your help patrick :)
Did we have to reduce the matrix to RREF or could we have left it at REF?
What induce you to convert last row to zero, if you lefy non zero elements there, then solution could be trivial, what you say?
Because you are not supposed to leave the numbers like that when they can be reduced to zero maybe?
otherwise its not row echelon form
Hey quick question, can a homogeneous system with a trivial solution also have a nontrivial solution? assuming there are no free varibles
great... thank you...!
thx a lot that helped..
how to know when to reduce it to row echelon form or reduced echelon form?
thanks
Hey Patrick tell me if an unknown constant is given such as lambda in the fourth equation then how to solve it
well if lambda is an unknown constant, just combine whatever it was previously into one unknown constant. such as if the last line is b(lamda)x1 + c(lambda)x2 = 0 then set b(lambda) to a value, say r. do everything like said in the video and lambda would equal r/b. hope that helps.
thanks dude
Impressive 👍👍👍👍👍👍
Thanks :)
At 7:17 do you always make x4 equal to a constant or does it depend on the matrix? Thanks
The variable you make equal to the constant/parameter depends on the number of *non pivot variables* in the matrix (also called the FREE variable, as mentioned in 7:00). That is, all the variables that are NOT associated with a pivot position will be made equal to the constant.
This of course means that in some cases, you can get situations where there are multiple parameters (i.e non pivot or free variables) from the matrix when writing out all possible solutions.
Note that when we are finding the linear independence of a set of vectors, this situation where we set a variable equal to a constant (e.g x4 = k) ONLY occurs in a *dependent set* of vectors.
Awesome!
Hey I have got a problem, how did you notice that the first can be changed into that? Could someone please explain why? And also "If a linear system has four equations and seven variables, then it must have infinitely many solutions." Is this true or false? Can you exlpain why?
when dealing with homogeneous systems, do you have to put it into rref? can't be do the same w ref?
Essentially, yes we do have to calculate rref, unless you choose a different method to solve for each variable. We need the solution to determine if it is trivial or nontrivial one. Note: When there is a free variable (like in this example), you cannot, by definition, use elementary row operations (or any other operation) to calculate the final rref. By definition, Reduced Row Echelon Form, must have a leading entry (1) as the only non-zero entry in its column. Knowing the definition of Echelon and Row Reduced Echelon form is important (esp. if you are studying linear algebra).
I appreciate sir
first we have to convert it into echelon form or reduced echelon form?
This is late but reduced usually makes it easier and if you have a ti-84 calculator (not sure if the older models do that as well) you can have it solve that matrix for you
since we have the last row = Zeros .. then we can say " inf. many Solutions " , right ?
thanks😊😊😊😊😊
In this semester , we took four subjects in our math : PDE , Matrices, Complex and Laplace transformation .
We literally die 😭
thank u sir
@xmarin7x ha : )
u deserve 100m
"You can pick your fav value for K, I'm going to pick +1 to make the arithmetic easy" - y you no pick 0 and make it even easier? XD
You can teach math even who has a lower than 70 IQ
How do you know which variable is the "free variable"?
variable that is not the leading term of any line, has no relationship to the others, hence that free variable is assigned a free parameter like, when solving you would write: Let x1 = s1, x2 = s2, xn = sn, where s is an arbitrary letter and n is the number of free variables. then you backtrack to solve.
What if the number of variables is greater than the rank of the matrix by 2 (Number of nonzero rows in the echelon form ) in this case I will be having 2 free variables how Can I solve the system in this case?
+Amr Wael let one variable equal to some arbitrary constant and then find the other variable in terms of this constant
What does Choosing the free variable stand for?
you put your pivot variablle in terms of the free variable. The free variable stands for any number
This video just saved my ass.
but what if we pick k=0? wouldn’t that turns out to be trivial too?
is k starts from zero?i-e k=o is possible???
why did not you plug k=0 ?
long last this video
making math easy
thanks
doing my best!
But if k=0 then wouldn't it be trivial?
How do you decide if it's nontrivial or trivial by inspection only? You have a system, that's it.
(im here jut for fun)but this looks like matrices on steriods!
is he doing reduced row echelon form?
+TheSummarizer Yes, he is. It's reduced because the pivot entries are 1, and the other entries in the pivot columns are 0. In (row) echelon form the pivot entries can be any number not equal to 0, and only the numbers below the pivot entries have to be 0.
Left hand
nice
Nice
is all 0 be one of the solution for this question?
yes! a trivial solution (all zeros) is always a solution to a homogeneous system of equations.
umm how that 9 become -9 ? sorry,, in the x2 = -9k ??
+Adrian Bullecer x number 2 =-9 not 2x
If one of the equations can be eliminated by row reduction, then the system of equations was not linearly independent.