Linear Algebra 1.2.2 Solution Sets and Free Variables

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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 :)

existentiol
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Honestly, Thank you so much. Felt so lost but this video along with your other ones are helping step by step.

rahulrathod
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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.

jonahp
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Found your videos much better than 3blue1brown! Please clarify more on vector spaces and Linear transformation

thatdecemberguy
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yippee! still following....good teacher step by step! Feeling pumped!

ticketpirates
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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

mahmoudalbahar
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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

opufy
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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

liamhoward
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the handwriting form is way better than the ppt. very cool!

xihengli
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thanks for your wonderful video, that's so helpful

laixu-mk
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thank you for this lesson that let me recall what i learned 25 years ago

chuckc
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Your videos are great. What textbook do you use in the linear algebra class?

denniscarroll
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thanks.. helpful ! <3 love from Pakistan

ahsanhaq
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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!

edwardjones
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3:52 shouldn't x2= 4+(-6) = -2. Therefore, the solution would be (-29, -2, -6)

shanikirson
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why would we want to write an augemented matrix instead of a coefficent matrix?

gewrgia
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So, any time we wind up with at least one free variable, we will have an infinite number of solutions?

gregdoty
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0:59 How is this matrix not in reduced row echelon form? It is, right?

basdewaal
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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?

rachelprojects
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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

JuliaZhang-qbyy