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Convert following matrix into Reduced Echlon Form:
just reduce echelon form right?
so what we need is an upper triangular matrix which means that below the main diagonal we need all 0's we can achieve this by using row operations by swapping rows, multiplying a row by a scalar number, or add/subtract rows and replace the old row with the new result after that computation
keep in mind that we need a 1 always at the top left the 1 also has to go towards the right and the bottom... we also need a row of zeros
wait I've solve this, can you please tell me is this right?
|dw:1438863440670:dw| so the places I've circled is the main diagonal so we need to get rid of the 3 , -1, and 4 using row operations... yeah sure.
anyway... -3r1 + r2 -> r2 r1+ r3 - > r3
hmmm gonna make life a bit easier by 1/2r3->r3
1/10 r3 - >r3 1/11 r2 - > r2
do you have a matrix that looks like this? ^
how did you get 20/11 ?
by multiplying Row 2 by -2 and add it in row 3(-2*1/11 + 2)
hmmm dealing with fractions that early is killer though
because what I've done to the third row was divide by 1/2 all over row 3 so I would have just 0 1 1
so when I had 0 11 1 0 1 1 I distributed the negative sign on row 2 and then multiply row 3 with 11 -r2+11r^3 -> r^3
Uhm so i i did wrong?
0 -11 -1 0 11 11 adding them together would make it 0 0 10 and then 1/10r3 ->r3 to get rid of that 10 to have 0 0 1
but in last it becomes in reduced echlon form; isnt?
and then to have the 1 in the middle I had to divide the whole second row by 1/11
somehow our first rows are different but the second and third is the same..
does my answer is correct?
it should be as long as it followed the reduce echelon rules.. and it does. we do have an upper triangular matrix and our 1 is going down and shifted to the right
Yes! So our answers are different, but rules are same!
hmm I'm assuming that an extra step was taken to get rid of the -2 on the first row. unless we are required to get RREF form, that's not required. What's required is the numbers until the main diagonal go to 0.
check this one out... it also did something different but ended up with the same thing before taking RREF form oh that 1 in the third row got blocked out, but noticed how this calculator did something I disagree with, but still ends up following the rules?
wait let me check it
that was done by a calculator bot... I really hated that 0 0 54/11 but by multiplying 11/54r3 -> r3 0 0 1
ugh a typo.. yeah this is way late at night... on my end.
still has that big fraction 20/11 on row 3 but 11/20r3 - >r3 makes it 0 0 1
my answer is correct!!!
XD you're welcome... these questions always come up very late at night, so typos are bound to happen... typed too fast XD
Its okay, when i post i dont run xD
yay I got it right too XD |dw:1438865533703:dw| row operation #5 on the correct image has the same matrix I had... only in fewer steps.
but as long as the rules aren't broken for row echelon form, then everything will be fine ^_^
There is still a confusion uhm the answer you showed me are correct one but in my answer in fist row there is -9/11, still is it right?
you mean for this right http://assets.openstudy.com/updates/attachments/55c34eede4b06d80932e63a6-imtiaz7-1438864107356-20150806_172006.jpg ?
as long as you have an upper triangular matrix which means all 0's below the main diagonal and have a 1 on the top left that goes to the bottom and right... I don't see a problem.
Once again thanks a lot :)
we could RREF that last matrix with that -9/11 if you want
to show that we can get an identity matrix out of some more row operations..
|dw:1438866425233:dw| this was the matrix you had in REF form right?
I would use row 3 to get rid of the remaining numbers above the main diagonal, but I hate dealing with fractions so 11r2 - >r2 11r1 -> r1
-r3+r2 - > r2 -9r3 + r1 - > r1
1/11r2 - >r2 1/11 r1 -> r1... I wish we can give more than 1 medal XD
|dw:1438866634762:dw| a-ha identity matrix! so we got RREF XD
which means your row operations from earlier were correct... otherwise we wouldn't get RREF