Matrix question (1 Viewer)

[ND]

I've got this question in an assignment and i'm not exactly sure what it asks:

Suppose R andS are 2x3 row-reduced echelon matrices and that the linear systems Rx=0 and Sx=0 have exactly the same solutions. Prove htat R=S.

I'm not sure what Rx=0 and Sx=0 means. Does that mean that they are just homogenous matrices? Or is x a constant? (though i really don't thinhk they'd make x a constant)

Any help would be greatl appreciated (oh and it'sdue wednesday, so please help soon).

 

[McLake]

Isn't "x" a 1xn matrix? I could be wrong ...

 

[xiao1985]

Isn't "x" a 1xn matrix? I could be wrong ...

i partially agree with mclake... x should be just a n x 1 vector (was it row or column??? can't remember)

edit:
*flips thru linear algebra notes...

maybe use the different pronumerals in matrix R and S, then multiply it out... it should work out... i try to work 2morrow or day after.. right now too tired @@ =p

 

[Affinity]

x could be any 3*k matrix.
the 0 corrosponds to the 2*k zero matrix.

It's just a tedious case bash I suppose..

 

[jm1234567890]

what does hat mean again?

i don't remember seing any matrixes with hats?

 

[Affinity]

as vector with a hat is a unit vector... don't know about matricies

 

[ND]

Thanks for the help guys. I did it, but i'm not sure if it's a little dodgy. Can someone plz check it?

Since they're row echelon matrices, they can be expressed:

R=[1,0;0,1;r1,r2], S=[1,0;0,1;s1,s2]

Now upon multiplying by column vector x=[a,b,c] you get:

Rx=[a+c*r1,b+c*r2], Sx=[a+c*s1,b+c*s2]

But since they have the same solutions (as given in the question), we can equate the columns [this is the part i'm not entirely sure about) then cancel, showing that r1=s1 and r2=s2.

 

[jm1234567890]

as vector with a hat is a unit vector... don't know about matricies

" Prove htat R=S."

lol, i though that said hat R = S

 

[evilc]

" Prove htat R=S."

lol, i though that said hat R = S

thats exactly what i read it as

 

[+Po1ntDeXt3r+]

hmm i got a matrix question too.. grrr.. cant figure ND ones out either
im not sure why im stuck.. need help with method..

2.Let u = (2 ,-5 ,-1); v = (3,-1,5); w = (6,-2, 3); x = (-7,-4,6)

e)Do some or all of the vectors u; v; w; x form a basis for R3? If so give a basis for R3 in terms of them?

 

[wogboy]

hmm i got a matrix question too.. grrr.. cant figure ND ones out either
im not sure why im stuck.. need help with method..

2.Let u = (2 ,-5 ,-1); v = (3,-1,5); w = (6,-2, 3); x = (-7,-4,6)

e)Do some or all of the vectors u; v; w; x form a basis for R3? If so give a basis for R3 in terms of them?

Let the matrix A be the matrix with columns u, v, w, and x. Do Gaussian elimination to transform that matrix into row echelon form. Then circle each element in that row reduced matrix which is the first non-zero element on it's row of the matrix (these are pivot elements). Any columns you see in this matrix with pivot elements is a pivot column. Now go back to the original matrix A (the one BEFORE you row reduced it), and each column in that matrix corresponding to a pivot column is an element of a basis for R3 (e.g. if the 1st, 2nd, and 4th columns of the row reduced matrix is a pivot column then the 1st, 2nd, and 4th columns of the ORIGINAL non row reduced matrix form a basis for R3)

Note: The dimension of R3 is 3, so there MUST be precisely 3 vectors in its basis (so obviously all four of those vectors can't form a basis for R3). Also note that there's infinitely many bases for vector spaces like R3 so don't worry if your answer doesn't match the answer at the back of the book.

 

[KeypadSDM]

Rx = 0
Sx = 0

Rx - Sx = 0
(R - S) x = 0

But x are non-sero soultions?

Thus R - S = 0 (Where 0 is the 0 2 * 3 matrix)
:. R = S

Ok, now why is that wrong?

 

[KeypadSDM]

Rx = 0
Sx = 0

Rx - Sx = 0
(R - S) x = 0

But x are non-sero soultions?

Thus R - S = 0 In order that x is a solution to that set of linear equations.(Where 0 is the 0 2 * 3 matrix)
:. R = S

Ok, now why is that wrong?