Getting ready for linear algebra next semester by binging this entire playlist and taking notes (the teacher is apparently less than coherent, so I'm trying to get ahead)... these videos are great! Very clear, straightforward, and easy to follow. Thanks for this! I'll pass this along to my classmates if any of them are struggling :)
I want to emphasize a point that only one other commenter has mentioned and it is that at 0:59, the matrix depicted is actually in reduced row echelon form (RREF). You mention that because of the -5 and 1, the matrix cannot be in RREF however, the two values are not in pivot columns. The pivots columns of x1 in row 1 and x2 in row 2 do not have any non zero values in them. Also since all other conditions are met, this matrix is in RREF. Just wanted to point that out.
she basically mean that it's not in the form but she didn't mean it can't be a RREF it can be but at that situation it's not watch the video before it you will get what i'm saying...
Summarizing this video : 1. Pivot element : The first non-zero element of each row when matrix is in echelon form is called pivot element 2. Pivot column : column containing pivot element is called pivot column 3. Basic and free variable : Every variable has its column. if that column is pivot then that variable is basic otherwise it is free variable. 4. If a system has at least one free variable then it has infinite many solutions(Consistent).
please answer these two questions: why non pivot columns are free variables? what is the proof of it? I love understanding the rule... not just saving it without understanding. and thanks for your very good videos
I am not sure I understand your question. You only get free variable is pivot value is 0. It's simply saying, we can get the value of other variables by using it as dependent on free variable. (think it as function).
what if it is a 4x3 matrix and works out where the last row is all 0s and but the other 3 rows have a solution. [[1,0,0:1],[0,1,0:2],[0,0,1:3],[0,0,0:0]]. Is this a unique solution or infinitely many? is the last equation just redundant. Thanks for the video!
12:00 does x_2 necessarily HAVE to be the free variable? Because couldn't we technically express the solution in terms of x_1 as well: (x_1 = x_1, x_2 = - (1/3)x_1 - (5/3), x_3 = 3)? In this case x_1 is free and x_2 bounded, but isn't this a valid solution as well?
Also, how did you know that X2 is a free variable? Assume the last column turns out to be 0 1 0 3, does that mean that X3 is the free variable? Is there a way to prove it?
Getting ready for linear algebra next semester by binging this entire playlist and taking notes (the teacher is apparently less than coherent, so I'm trying to get ahead)... these videos are great! Very clear, straightforward, and easy to follow. Thanks for this! I'll pass this along to my classmates if any of them are struggling :)
I wish you were my actual prof.
You can transfer to Bellevue University!
Honestly, Thank you so much. Felt so lost but this video along with your other ones are helping step by step.
Thanks! Glad they were helpful!
I want to emphasize a point that only one other commenter has mentioned and it is that at 0:59, the matrix depicted is actually in reduced row echelon form (RREF). You mention that because of the -5 and 1, the matrix cannot be in RREF however, the two values are not in pivot columns. The pivots columns of x1 in row 1 and x2 in row 2 do not have any non zero values in them. Also since all other conditions are met, this matrix is in RREF. Just wanted to point that out.
she basically mean that it's not in the form but she didn't mean it can't be a RREF it can be but at that situation it's not watch the video before it you will get what i'm saying...
Summarizing this video :
1. Pivot element : The first non-zero element of each row when matrix is in echelon form is called pivot element
2. Pivot column : column containing pivot element is called pivot column
3. Basic and free variable : Every variable has its column. if that column is pivot then that variable is basic otherwise it is free variable.
4. If a system has at least one free variable then it has infinite many solutions(Consistent).
Found your videos much better than 3blue1brown! Please clarify more on vector spaces and Linear transformation
I have videos on each of those topics. Here is the entire playlist: ua-cam.com/play/PLl-gb0E4MII03hiCrZa7YqxUMEeEPmZqK.html
best teacher everr!!!!!!
yippee! still following....good teacher step by step! Feeling pumped!
Thanks!
thank you for this lesson that let me recall what i learned 25 years ago
12:45 I just use the matrix to get to Reduced echelon form instead of plugging in 3. So its row 1 subtracted by 4 x row2
You are saving my life😭
thanks for your wonderful video, that's so helpful
So their are 2 ways a variable can be free then correct?
1. A column has no pivot
2. A row has all zero values
the handwriting form is way better than the ppt. very cool!
0:59 How is this matrix not in reduced row echelon form? It is, right?
that's what i was thinking as well
watch the video before you will understand
the values above the pivot 1s have to be 0
WHat teh hell bra!
Now I understand this thing of free variables...
please answer these two questions:
why non pivot columns are free variables?
what is the proof of it?
I love understanding the rule... not just saving it without understanding.
and thanks for your very good videos
I am not sure I understand your question. You only get free variable is pivot value is 0. It's simply saying, we can get the value of other variables by using it as dependent on free variable. (think it as function).
thanks.. helpful !
Thank you!
Your videos are great. What textbook do you use in the linear algebra class?
I use David Lay Linear Algebra and Its Applications 5th edition
@@SawFinMath thank you so much!
3:52 shouldn't x2= 4+(-6) = -2. Therefore, the solution would be (-29,-2,-6)
naw
here before this blows up
Hoping it does!
why would we want to write an augemented matrix instead of a coefficent matrix?
When you have to solve for the value of the variable, you augment with the constants in the system of equations.
@@SawFinMath thank you
what if it is a 4x3 matrix and works out where the last row is all 0s and but the other 3 rows have a solution. [[1,0,0:1],[0,1,0:2],[0,0,1:3],[0,0,0:0]]. Is this a unique solution or infinitely many? is the last equation just redundant. Thanks for the video!
12:00 does x_2 necessarily HAVE to be the free variable? Because couldn't we technically express the solution in terms of x_1 as well:
(x_1 = x_1, x_2 = - (1/3)x_1 - (5/3), x_3 = 3)? In this case x_1 is free and x_2 bounded, but isn't this a valid solution as well?
You can solve it that way but technically x_2 is the only free variable.
So, any time we wind up with at least one free variable, we will have an infinite number of solutions?
Yup. Because you will have a solution for every value of your free variable.
Kimberly brehm 🔥🔥
Really helpful! Thanks :)
Glad it was helpful!
Why is it possible to solve 3 three-variable system with only two equations using matrices? I'm referring to the last problem in this video
Also, how did you know that X2 is a free variable? Assume the last column turns out to be 0 1 0 3, does that mean that X3 is the free variable? Is there a way to prove it?
Thank you . what is the textbook of this course?
The classic. Lay's 5th edition.
to clarify at around the 13 min mark you started writing x_2 as x^2 was that just a mistake or am i missing something?
nevermind you corrected in the video
Can I get passed my whole semester by completing this playlist?
Complete
1:19 wtf