MAth133 assignment (1 Viewer)

[Dash]

any1 dun tha bloody assignent...

im full strugglin/....

need help with question 6 and 7...

wat do we have to do with question 6?? dun have a clue??

and question 7 how u prove it??

 

[...]

man, if your friend is doing 133, she/he shouldn't be struggling in assignment questions

 

[flyin']

6. Describe explicitly all possible 2 × 2 row-reduced echelon matrices. (It's hard to restate the question further, I guess.)

(0,0;0,0), (1,0;0,0), (1,0;1,0) where (a,b;c,d) represents a 2x2 matrix with (a,b) the first column vector and (c,d) the second vector. Not sure whether (0,0;0,0) counts but sure the other two do.

7. Suppose R and S are 2×3 row-reduced echelon matrices and that the linear systems Rx = 0 and Sx = 0 have exactly the same solutions. Prove that R = S.

I think x is a 3x1 matrix.

Let R = (a,b;c,d;e,f), S = (g,h;i,j;k;l) and x = (p;q;r). Then show that "Rx = 0 and Sx = 0 have exactly the same solutions (ie. (p;q;r))" if and only if a=g, b=h, etc.

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Disclaimer - I might be making all this up!

 

[Dash]

hey flyin are you a 2nd year..

and how sure if ur answers are correct??

 

[flyin']

Yeah, I'm in second year. I've done 133 (last year, second sem) and 235 (this year, first sem).

How sure am I? I pretty sure. But then I did chuck a disclaimer at the end...

 

[RIZAL]

6. (0,0;0,0), (1,0;0,0), (1,0;1,0) AND (1, n, 0, 0) for all n is real.

7. This one is easy to visualise but hard to put on paper.

 

[Dash]

math 133 book

would any1 be interested in selling the math133 textbook second hand???

algebra and calculus??

 

[flyin']

"6. (0,0;0,0), (1,0;0,0), (1,0;1,0) AND (1, n, 0, 0) for all n is real."
Isaac, I don't think (1, n, 0, 0) is reduced-row echelon.

Dash, I'm trying to sell {Elementary Linear Algebra (Anton), Eighth Edition} for a friend. Asking price is ~$60.

 

[RIZAL]

"6. (0,0;0,0), (1,0;0,0), (1,0;1,0) AND (1, n, 0, 0) for all n is real."
Isaac, I don't think (1, n, 0, 0) is reduced-row echelon.

it is. see the definition of reduced-row echelon.