Question

Linear algebra question.... see attachment (part d)

Answer 1

I did the reduced row echelon form and it gives that all variables are equal to zero. Is it reasonable to say that if a system has a solution, and its augmented matrix is one of the form that the augmented part is 0, then the solution will have all variables equal to 0?

Answer 2

I know this is called the "trivial solution" but is there a case of an augmented matrix with the augmented part as zeros, where the solution will not be the trivial one?

Answer 3

hiii

Answer 4

Hello

Answer 5

i guess you know how to calculate the rank

Answer 6

when the system is homogeneous, the rank of coefficient matrix and the augmented matrix are equal.. so we may only perform operations on the coefficient matrix

Answer 7

I don't know what you mean by "rank"

Answer 8

calculate the rank of the coefficient matrix by converting it into row echelon form

Answer 9

You know row echelon form??

Answer 10

@chrisplusian

Answer 11

the solution to the set is the origin ...

Answer 12

3 planes can meet in a number of different ways for a graphical representation.

in this case they make a tepee, whose apes is 0,0,0

Answer 13

apex ... thats what i get for trying to be lyrical

Answer 14

Sorry I lost internet connection @jango_IN_DTOWN. I will show you what I did....

Answer 15

@amistre64 is there a case where you can have the augmented matrix, with the "augmented portion" (<---- not sure if this is correct terminology) equals zero, but the solution to the system is non-zero?

Answer 16

This is right..

Answer 17

if the matrix is full-rank (its determinant is not zero) , then only the trivial solution will give all zeros.
on the other hand if the matrix is not full rank, yes you can have non-trivial solutions

Answer 18

I have no idea what "rank" is.

Answer 19

That is not a topic we have covered yet

Answer 20

if you think of each row in the matrix as an equation,
then "rank" is the number of "independent" equations
if you have not learned this, you eventually will.

Answer 21

My question was.... if there is an augmented matrix with the augmented portion (the part to the right of the line) that is zero can there be another solution other than the origin? Like can there be a numerical solution? Or a dependant solution? I think I understand geometrically what the ideas are. If I end up with a numerical solution then there is a place that in R^3 the planes intersect and the solution is that point. If I end up with a consistent solution that is dependent than I have a line that at which all three planes intersect, and if I end up with an inconsistent system that implies that the three plans have no mutual point/s of intersection. What I can't decide is if having an augmented matrix with the augmented portion as all zeros actually tells me anything?

Answer 22

@phi we just discussed the idea of "lead variables" and "free variables", could this somehow be the same idea as "rank"?

Answer 23

@chrisplusian you know minors and cofactors of a matrix right? Then it will be very easy to define rank

Answer 24

I understand minors and cofactors :) That was what we discussed in our last class meeting.

Answer 25

ok they will define rank in the near classes I am sure.. Ok rank is "the greatest potisive integer r such that that matrix A has at least one non zero minor of order r".

Answer 26

Ok so maybe I am jumping ahead?

Answer 27

if you know of spans, then the matrix here spans R^3, the vectors are independant.

also, each equation defines a plane in R^3 such that they all meet at one point. only one point is common to all 3 planes.

Answer 28

Are there any things I should know about an augmented matrix with the augmented part as all zeros, that does not require the concept of "rank"?

Answer 29

@amistre64 what do you mean by spans?

Answer 30

if you have free variables, then you will not have full rank
if the number of lead variables is equal to the number of rows (equations) then you have full rank. (rank is the number of lead variables)