I’m really really really thankful for you, you have just made you print in this world and helped a lot of people like me, I’m grateful that there are people like you in this world
Is 'parametric vector form' always written as a span? I've seen people call the solution given at 4:30 as being in parametric vector form, but maybe I'm not understanding it correctly.
The solution set to a homogeneous system *can* always be written as a span, but it doesn't have to be. Solutions sets to other systems are covered in the next lecture, and are not able to be written as a span. I also talk about parametric vector solutions in the next lecture.
@@HamblinMath as we know span is of column vectors of A then why u take it as of x vector.. is it true. ..... And thnks to video these are really helpful.. Plz reply to ques sir 🙏🙏🙏🙏🙏
you're the goat bro. I understood finding the solutions for the homogeneous system of linear equations, but was confused on representing the solution set. 50 videos deep and yours was the light at the end of the tunnel 🆙🥹
Because there are no pivots in the 2nd and 3rd columns. I recommend watching Lecture 5 for more information (th-cam.com/video/kDbBTFvQgig/w-d-xo.html).
How about when its the other way around tho? When youre given the solution of a homogenous system and youre asked for the coefficient matrix, aka what the system is? I couldnt find it anywhere so any help would be greatly appreciated, ty
Even a seemingly simple equation like x = 0 in R^3 can be 'solved' in parametric vector form by using an augmented matrix, following your method at 7:00 (and next video). The equation x = 0 in R^3 is equivalent to the equation 1x + 0y + 0z = 0 , which gives us the augmented matrix [ 1 0 0 0 ] , which happens to be already in rref form ; it has one pivot term and 2 free variables. Then a little rewriting give us (x,y,z) = (0, y, z ) = y * (0, 1 ,0 ) + z * (0, 0, 1). We can let y and z be parameters, so the plane x = 0 in R^3 is equivalent to the set of vectors { s * (0,1,0) + t * (0,0,1), s∈R , t ∈ R} . That makes sense since the graph of x = 0 in R^3 is the yz plane, which is spanned by the vectors (0,1,0) and (0,0,1). This may seem somewhat pedantic, but it illustrates the point that free variables in the rref matrix do not technically require nonzero terms (however a pivot variable , also called basic or leading variable, must have a nonzero entry in rref form). Similarly we can 'solve' the plane y = 0 in R^3 which has rref form [ 0 1 0 0 ] gives us the vector solution set { s * ( 1, 0 , 0 ) + t * ( 0, 0, 1 ) | s, t ∈ R } . And the plane z = 0 in R^3 which has rref [ 0 0 1 0 ], gives us the vector solution set { s * ( 1, 0 , 0 ) + t * ( 0, 1, 0 ) | s, t ∈ R } . For the penultimate free variable case in R^3, the equation 0x + 0y + 0z = 0 corresponds to the augmented matrix [ 0 0 0 0 ] , which is already in rref form, and all variables are free. Since we can express (x,y,z) = x * (1,0,0) + y*(0,1,0) + z*(0,0,1) we can express the set of vector solutions { (x,y,z) } as { s * ( 1,0,0) + t * (0, 1, 0 ) + u * ( 0, 0 , 1) | s, t, u ∈R } . This also tells us R^3 is spanned by (1,0,0) , (0,1,0) and (0,0,1). Apologies for the length of this comment.
I have doubt, as you said "Homogeneous means "all of the same kind" here linear equation set to zero" My doubt is " why can't have all linear equation set to only one constant term say for e.g.,"b" instead of zero"?? Is that be homogeneous or not?
The term "homogeneous" is used in different ways in different mathematical areas. In linear algebra, it specifically refers to the case where the right-hand side of each equation is zero.
At 3:20 could someone please explain what he meant about x_3 being a free variable since it's not a pivot column? I'm confused. Any help would be great.
It means that, in the solution to the corresponding system of equations, x_3 can be any value, and then the values of the other variables will be determined based on the value of x_3. We say x_3 is "free" to be whatever value we want it to be.
The solution sets for *all* linear systems fall into one of three categories: (1) no solutions, (2) one unique solutions, (3) infinitely many solutions. A homogeneous system always has at least the solution where all of the variables equal zero, so possibility (1) can't happen for these types of systems. If there are *any* nontrivial solutions, then we must be in category (3), which means we have at least one free variable and infinitely many solutions. I hope this helps you understand this better!
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I’m really really really thankful for you, you have just made you print in this world and helped a lot of people like me, I’m grateful that there are people like you in this world
Wow, thank you so much. I'm glad my videos have helped you.
These videos are great! In depth and concise without being overly simple.
Thanks!
Mayn you are life saver
Edited: Had no idea about linear algebra the night before my exam but after watching his videos I scored 16/30 : ))
It helped me at the end moment when nothing was working out
Thank you so much❤️❤️❤️❤️❤️
Can we get them in one form like written or your book in the same order??
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Is 'parametric vector form' always written as a span? I've seen people call the solution given at 4:30 as being in parametric vector form, but maybe I'm not understanding it correctly.
The solution set to a homogeneous system *can* always be written as a span, but it doesn't have to be. Solutions sets to other systems are covered in the next lecture, and are not able to be written as a span. I also talk about parametric vector solutions in the next lecture.
@@HamblinMath as we know span is of column vectors of A then why u take it as of x vector.. is it true. .....
And thnks to video these are really helpful..
Plz reply to ques sir 🙏🙏🙏🙏🙏
you're the goat bro. I understood finding the solutions for the homogeneous system of linear equations, but was confused on representing the solution set. 50 videos deep and yours was the light at the end of the tunnel 🆙🥹
thank you x 100 million times!!!!!!
thanks so much for such a helpful video!😄👍🏻
Great job on these videos. If you create a Patron account, I will be your first patron.
I appreciate that, but I make these videos as a resource for my students. They will always be free!
at 7:35 how does he know that y and z are the free variables?
Because there are no pivots in the 2nd and 3rd columns. I recommend watching Lecture 5 for more information (th-cam.com/video/kDbBTFvQgig/w-d-xo.html).
@@HamblinMath I see it now, thank you so much for these videos!
How about when its the other way around tho? When youre given the solution of a homogenous system and youre asked for the coefficient matrix, aka what the system is?
I couldnt find it anywhere so any help would be greatly appreciated, ty
did u find any video on it ? if so, can u pls share..
@@wajeehakhalid4496 well I have graduated now I don't remember if I did XD gl tho mate
Very helpful thank you.
Even a seemingly simple equation like x = 0 in R^3 can be 'solved' in parametric vector form by using an augmented matrix, following your method at 7:00 (and next video).
The equation x = 0 in R^3 is equivalent to the equation 1x + 0y + 0z = 0 , which gives us the augmented matrix [ 1 0 0 0 ] , which happens to be already in rref form ; it has one pivot term and 2 free variables.
Then a little rewriting give us (x,y,z) = (0, y, z ) = y * (0, 1 ,0 ) + z * (0, 0, 1). We can let y and z be parameters, so the plane x = 0 in R^3 is equivalent to the set of vectors { s * (0,1,0) + t * (0,0,1), s∈R , t ∈ R} . That makes sense since the graph of x = 0 in R^3 is the yz plane, which is spanned by the vectors (0,1,0) and (0,0,1).
This may seem somewhat pedantic, but it illustrates the point that free variables in the rref matrix do not technically require nonzero terms (however a pivot variable , also called basic or leading variable, must have a nonzero entry in rref form).
Similarly we can 'solve' the plane y = 0 in R^3 which has rref form [ 0 1 0 0 ]
gives us the vector solution set { s * ( 1, 0 , 0 ) + t * ( 0, 0, 1 ) | s, t ∈ R } .
And the plane z = 0 in R^3 which has rref [ 0 0 1 0 ],
gives us the vector solution set { s * ( 1, 0 , 0 ) + t * ( 0, 1, 0 ) | s, t ∈ R } .
For the penultimate free variable case in R^3, the equation 0x + 0y + 0z = 0
corresponds to the augmented matrix [ 0 0 0 0 ] , which is already in rref form, and all variables are free. Since we can express (x,y,z) = x * (1,0,0) + y*(0,1,0) + z*(0,0,1) we can express the set of vector solutions { (x,y,z) } as { s * ( 1,0,0) + t * (0, 1, 0 ) + u * ( 0, 0 , 1) | s, t, u ∈R } .
This also tells us R^3 is spanned by (1,0,0) , (0,1,0) and (0,0,1).
Apologies for the length of this comment.
I have doubt, as you said "Homogeneous means "all of the same kind" here linear equation set to zero"
My doubt is " why can't have all linear equation set to only one constant term say for e.g.,"b" instead of zero"?? Is that be homogeneous or not?
The term "homogeneous" is used in different ways in different mathematical areas. In linear algebra, it specifically refers to the case where the right-hand side of each equation is zero.
@@HamblinMath Thank you for your reply.
At 3:20 could someone please explain what he meant about x_3 being a free variable since it's not a pivot column? I'm confused. Any help would be great.
It means that, in the solution to the corresponding system of equations, x_3 can be any value, and then the values of the other variables will be determined based on the value of x_3. We say x_3 is "free" to be whatever value we want it to be.
@@HamblinMath That makes sense. Thanks for the explanation. You've gained a subscriber!
I have a question
Why is it that all non trivial solutions have to have free variables ? Why cant it be without free variables?
The solution sets for *all* linear systems fall into one of three categories: (1) no solutions, (2) one unique solutions, (3) infinitely many solutions. A homogeneous system always has at least the solution where all of the variables equal zero, so possibility (1) can't happen for these types of systems. If there are *any* nontrivial solutions, then we must be in category (3), which means we have at least one free variable and infinitely many solutions.
I hope this helps you understand this better!
Nah you trippin 💯
Poor pedagogy
Would you care to elaborate?
I'm sorry but I really didn't understand what is parametric vector form.