Thanks so much for the feedback, Egemen! There's more content I have that isn't uploaded. So, if there's a topic that's troubling you, I may have a video on it. :-) Best of luck with your educational endeavors in the meantime!
Happy to clarify! Note that the solution space of any homogeneous linear system of R^n is a subspace of R^n. Let's call this subspace "W". Since R^n is finite dimensional, then so is "W". Putting these two facts together, we can conclude that W is a finite dimensional vector space. Therefore, it has a basis. :-)
can we say that basis are the solution that satisfies system by writing its parametric equation ? (what i mean basis generates parametrical solution which satisfies the system)
Happy to clarify! The vectors that result from defining the parametric equations can be used for the basis. They're not unique, though. For instance, and as an aside, any scalar multiple of these vectors can be used as basis vectors. :-) Hope this helps!
@@larrychoraszewski1125 Thats great thanks! For differential equations, i have a question. Can we tell that general solution is not unique by solving different methods / way ? What i mean general solution is unique for ODE ? Thanks really for your help aswell for ODE
No worries! Rows 2 & 3 are proportional to Row 1. That is, they're equivalent to the first equation. Geometrically, we would have "overlapping" planes, which is why this system has infinitely-many solutions. :-)
so in other words you could say that the dimension of a solution space of a system of linear equations (SLE) is definend by n (Number of variables) - rank (of the SLE) = number of free Variables..
Glad you asked! You would first find the general solution for the homogenous linear system Ax=0, as demonstrated in the video. Next, you'll find the specific solution to Ax=b and simply add this to the general solution from Ax=0 that you first found. :-) Let me know if this helps out. Thanks!
These two rows are proportional to R1; i.e., a constant multiple of R1. In other words, the equations in these two rows are equivalent to the equation in R1. :-)
Thank u bro literally saved me 3 hours before the midterm
Thank you for the support, and I'm happy to hear it helped you out on the midterm! 🙂
im not even from your university but you explained it so well in 1/10 of the time that my lecturer took to explain, thank you!!!!!
Thanks so much for the feedback, Egemen! There's more content I have that isn't uploaded. So, if there's a topic that's troubling you, I may have a video on it. :-) Best of luck with your educational endeavors in the meantime!
Bruh this is so helpful. When you actually try to understand this, its totally easy. Thank you
Glad to hear you found it helpful! And welcome to the Linear Alge-bruh fam!
@Ryan Choraszewski BRUH!
Thanks Professor❤
I'm glad you found it helpful, and thanks for the support! :-)
thankyou sir, it was helpful.🌻
So happy to hear this. Being helpful is what I strive for. 🙂Thank you for watching!
thank u so much
Any time! Thanks for the support, too! 🙂
Nice description, go ahead BRO
Many many thanks! :-)
But how can it even have a basis when it should not be able to form a basis since it fails to be linearly dependent (it forms a nontrivial solution)?
Happy to clarify! Note that the solution space of any homogeneous linear system of R^n is a subspace of R^n. Let's call this subspace "W". Since R^n is finite dimensional, then so is "W". Putting these two facts together, we can conclude that W is a finite dimensional vector space. Therefore, it has a basis. :-)
can we say that basis are the solution that satisfies system by writing its parametric equation ? (what i mean basis generates parametrical solution which satisfies the system)
Happy to clarify! The vectors that result from defining the parametric equations can be used for the basis. They're not unique, though. For instance, and as an aside, any scalar multiple of these vectors can be used as basis vectors. :-) Hope this helps!
@@larrychoraszewski1125 Thats great thanks! For differential equations, i have a question. Can we tell that general solution is not unique by solving different methods / way ? What i mean general solution is unique for ODE ? Thanks really for your help aswell for ODE
Newbie here, why is rref all zeros on Row 2 and 3? shouldn't all rref have diagonal 1's and rest are 0s?
No worries! Rows 2 & 3 are proportional to Row 1. That is, they're equivalent to the first equation. Geometrically, we would have "overlapping" planes, which is why this system has infinitely-many solutions. :-)
@@larrychoraszewski1125 Thank you :)
@@edizon204 No problem!
thank you nice explanation
Thanks Howard!
Thank you!
so in other words you could say that the dimension of a solution space of a system of linear equations (SLE) is definend by n (Number of variables) - rank (of the SLE) = number of free Variables..
Yes, at least for a homogeneous system of linear equations.
Thank you.
No problem! Thanks for watching!
r3 = all value not zero
You referring to the matrix prior to RREF?
THANK YOU
No problem! :-)
Thank you
No problem!
How would you do this for a non homogeneous system?
Glad you asked! You would first find the general solution for the homogenous linear system Ax=0, as demonstrated in the video. Next, you'll find the specific solution to Ax=b and simply add this to the general solution from Ax=0 that you first found. :-) Let me know if this helps out. Thanks!
Got it, thank you!
@@samr3100 Happy to help! 🙂
Very helpful lecture sir 😊
Thank you! Glad it helped! :-)
Why do R2 and R3 columns be of zeros in rref?
These two rows are proportional to R1; i.e., a constant multiple of R1. In other words, the equations in these two rows are equivalent to the equation in R1. :-)
Didn't understand
So sorry... :-(