Linear Algebra - Lecture 10 - Homogeneous Linear Systems

In this lecture, we define "homogeneous" linear systems, and discuss how to find the solutions to these systems in parametric vector form.

[ayubseeed]

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[soz]

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[taiebxr]

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Edited: Had no idea about linear algebra the night before my exam but after watching his videos I scored 16/30 : ))

[aniketchakraborty]

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[laurenpark]

thanks so much for such a helpful video!😄👍🏻

[usrakhan]

Can we get them in one form like written or your book in the same order??

[gaussianelimination]

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.

[pooyashoghi]

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[starliaghtsz]

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

[thejordansandsbs]

at 7:35 how does he know that y and z are the free variables?

[AJ-fohp]

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?

[amaliarios]

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.

[ws]

I have a question
Why is it that all non trivial solutions have to have free variables ? Why cant it be without free variables?

[maxpercer]

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.