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Linear algebra (basis)

This is the question:
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And this is the mark scheme but I’m confused because I didn’t get the same after doing Gaussian Elimination and can’t tell if it does span(solution exists) with my answer.
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(edited 1 year ago)
They look the same to me (I didn't bother checking the last ERO, too messy to care at this time of day). They just differ by a multiple, so really it doesn't make a difference.

It's like "x+2y=1" and "2x+4y=2" are the same.

EDIT: That said, the matrices certainly spans the vector space. Essentially to check span, what you want is to find c1, ..., c4 in terms of alpha, beta.... But it's sufficient to stop where the solution ends (why?).
(edited 1 year ago)
Original post by tonyiptony
They look the same to me (I didn't bother checking the last ERO, too messy to care at this time of day). They just differ by a multiple, so really it doesn't make a difference.

It's like "x+2y=1" and "2x+4y=2" are the same.

EDIT: That said, the matrices certainly spans the vector space. Essentially to check span, what you want is to find c1, ..., c4 in terms of alpha, beta.... But it's sufficient to stop where the solution ends (why?).


Yeah my workings are too messy to be honest 😅.So can I say a solution exists because it seems unique on the right hand side?
Original post by Stress2638,,,
Yeah my workings are too messy to be honest 😅.So can I say a solution exists because it seems unique on the right hand side?


Your working is fine (quite clear to read tbh, so well done). I'm just not bothered to check algebra beyond 5-seconds of mental arithmetic.

Remember uniqueness of solution of a system is determined on the left. The system could not have a solution if some row reads "0 times x = 1" or something like that.
Since you don't get a row of zeros after Gaussian Elimination, the system must have a unique solution. In fact you can write down what the solution is, but it's not necessary here.
(edited 1 year ago)
Okay that makes sense! 👍

and for this question (8aii) I need to find the basis for the null space of A.If I remember correctly there is two ways of doing this by checking if it’s linearly independent and the span(unless you know the dimension you don’t need to check both).I found the span and I got 0=gamma +2alpha which is inconsistent but in the marking scheme they only check if it’s linearly independent(and says it does form form a null space?)

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Never,Ind I figured I need to use the eq rank(A) +nullity(A)=m so I do need to put into row echelon form and see how many free parameters it has…
(edited 1 year ago)
Hmm... I'm not sure what you are trying to do here.
(I'm thinking in linear transformation here, because matrices and LT are practically the same thing)

Recall what the nullspace(/kernel) of a matrix(/LT) is: it is just the set of all the vectors that get mapped to the zero vector.

Not much use here, huh... Maybe I'm unaware of your "method".

EDIT: Also the system is not necessarily inconsistent. In fact it is consistent if and only if 2alpha+gamma = 0, which is possible. That said, I don't know why you do that.
(edited 1 year ago)
Original post by tonyiptony
Hmm... I'm not sure what you are trying to do here.
(I'm thinking in linear transformation here, because matrices and LT are practically the same thing)

Recall what the nullspace(/kernel) of a matrix(/LT) is: it is just the set of all the vectors that get mapped to the zero vector.

Not much use here, huh... Maybe I'm unaware of your "method".


My method didn’t work out in the end but I read through the notes and figured it out in the end lol.What I was talking about before wasn’t right and that was for working out the no.of columns of a.

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