Lec 7 | MIT 18.06 Linear Algebra, Spring 2005

Lecture 7: Solving Ax = 0: Pivot Variables, Special Solutions.

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Category:

Education

Tags:

• matrix theory
• linear algebra
• systems of equations
• vector spaces
• determinants
• eigenvalues
• similarity
• positive definite matrices
• least-squares approximations
• networks
• Fourier transforms
• Markov processes

[truongvoky012345]

very good

[gavilanch]

Awesome, i´m glad I found this videos =D

[jojoejoe]

@shinyralle

When you switch the columns of R matrix, you need to switch rows of x vector to make sure it matches original linear equations.

For example,

R11(x1) + R12(x2) = 0

R21(x1) + R12(x2) = 0

in Matrix form,

|R11 R12||x1| equals |R12 R11||x2|

|R21 R22||x2| |R22 R21||x1|

[SunnyAudiegk409]

soso ich bin blond

[scotchandsoda123]

@shinyralle all he wants is, that you get a better view for that what he wants you to see. that why he writes it that way

[thebigfootme]

excellent!!!

[shinyralle]

at 26:13 I dont get how he can switch column 2 and column 3 to get the identity matrix in the first block of [ I F] ? You cant chage the order of pivot columns just like that ? PLEASE answer this for me someone! 1love

[akhil089]

thanks professor strang

[RadekLudva]

@freakiest421 "there is one special vector for every free variable"

[freakiest421]

linear combination of just two special vector will produce only 2D vector space. How do you know that's the complete solution?

[kotofu]

i just got a rather low grade in my class, wish i spent more time watching his videos...