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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
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?
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

hiii
Hello
i guess you know how to calculate the rank
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
I don't know what you mean by "rank"
calculate the rank of the coefficient matrix by converting it into row echelon form
You know row echelon form??
the solution to the set is the origin ...
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
apex ... thats what i get for trying to be lyrical
Sorry I lost internet connection @jango_IN_DTOWN. I will show you what I did....
1 Attachment
@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?
This is right..
  • phi
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
I have no idea what "rank" is.
That is not a topic we have covered yet
  • phi
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.
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?
@phi we just discussed the idea of "lead variables" and "free variables", could this somehow be the same idea as "rank"?
@chrisplusian you know minors and cofactors of a matrix right? Then it will be very easy to define rank
I understand minors and cofactors :) That was what we discussed in our last class meeting.
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".
Ok so maybe I am jumping ahead?
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. |dw:1444587665421:dw|
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"?
@amistre64 what do you mean by spans?
  • phi
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)
a span is a set of vectors that allows us to reach all the points in a given space. for R^2 (1,0) and (0,1) are the standard vectors that allow us to span the enirety of a plane. a basis, is a spanning set of vectors that are independant (none can be defined as a linear combination of the others)
  • phi
You could take a peek at the later chapters in your book
if Ax=0 only has the trivial solution .. then A forms a basis, i beleive
@phi my professor did not elaborate on the free variables, or lead variables yet. It was just a side note that when you are solving a system, any row that has a non-zero value after you have put it in reduced row echelon form, is considered to have a lead variable. If a column does not have a lead variable then it is a free variable. If you have a dependent system, solve for all the free variables and then write it in vector form.
  • phi
yes, and if you have a dependent system, then expect there to be non-trivial solutions to Ax=0
  • phi
But I am sure you will see soon in your course.
So to summarize if you have the augmented portion as zeros, then you will either have the trivial solution or a dependent system?
  • phi
yes there are other implications: determinant of the dep. sys is 0 inverse of A does not exist
As far as a numerical solution, a single point, if the augmented portion is all zeros (before the reduced row echelon form) then can you conclude that there will be no single point other than the origin that would be a solution to the system?
no
solution to 3 planes are either a point, a line, a plane, or nothing
  • phi
you can use row reduction to help decide that. Otherwise, only God knows
3 parallel planes have no common points 3 coincident planes are common at every point etc
Thank you all :)
spose our setup defines a line in space that does not pass thru the origin. like 3x + 4y -2 = 0 x=0 and y=0 is not a solution. if i understand your question correctly
No what I was asking about was in reference to an entire system.

Not the answer you are looking for?

Search for more explanations.

Ask your own question