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just a quick question and sorry for being messy =/
[ 1 2 3 ... n ]
[ n+1 n+2 n+3 ... 2n ]
[ 2n+1 2n+2 2n+3 ... 3n ]
[ . . ]
[ . . ]
[ . . ]
[(n^2 -n +1)(n^2 -n +2) (n^2 -n +3) ... n^2 ]
I am asked to find the row echelon form of the nxn matrix.
What i did is minused n from row 2 then minused 2n from row 3 and so on... untill the last row where i minused (n-1)n
this gave me an nxn matrix with identical rows containing 1 2 3 ... n
Would this be sufficent as the row echelon form of the matrix or simplify it further?
Thanks in advance for the help
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Hi samuel12;
Did you try your idea on a 4x4 matrix as a test?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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yeah i think i got a bit confused=/ let me get back to you haha
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Hi samuel12;
No problem.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Umm i'm not going to bother posting what i got unless anyone actually wants the answer if so speak up.
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Hi samuel12;
The answer! I don't even know what the question is. Better start with that!
I see what you are saying! It's for the nxn matrix that you want the rref of. Post what you have discovered, I will look at it. I can't guarantee I will understand it!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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the first row operation will be...
minus R1 from each of R2 through to Rn
then
divide all rows from R2 through to Rn by n
then
minus (n-1)R2 from each R3 to Rn (where n is the row number i.e. a constant)
now finally R2 - R1
to get....
[1 2 3 ... n]
[0 -1 -2 ... 1-n]
[0 0 0 ... 0]
[0 0 0 ... 0]
etc etc
which is the given nxn matrix in row echelon form (i think haha)
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Hi;
now finally R2 - R1
Do I store R2-R1 back into R2?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi easycalculation;
Some nice calculators there but no biorhythm one.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi;
now finally R2 - R1
Do I store R2-R1 back into R2?
Umm yeah so like R2---> R2 - R1
if that makes sense=)
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Hi;
No like R2-R1 -> R2
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Well yeah different notation in different countries i guess haha
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