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http://mypat.in/PaperImages/3/425/Q_4_004.gif

can you factor all of the elements

@UnkleRhaukus ,how exactly??

yeah maybe not, i really dont want to do it the normal way

Normal way-->never :D It would take ages :P

but it would work in theory, but i would make about ten mistakes

can we just do it the foolish way?

Hmm, tried C1 - C2?

I'm going to try this problem

Have no fear, for inky is here!

@shivam_bhalla I HAVE AN IDEA

aei+bfg+cdh-ceg-bdi-afh
=a(ei-fh)+b(fg-dj)+c(dg-eg)

b=d
f=h

c=g

aei+bfg+cdh-ceg-bdi-afh
aei+bf^2+cbf-ec^2-ah^2

I'm wikipediaing, and not making any progress.

LOL. @ "wikipediaing" . :)

wolfram!!!

How to wolfram this ??

factor out the commons!!!

@experimentX , how exactly ?? I mean row wise or column wise. And moreover, what to factor out ??

lol ... factoring out would lead to 3 more matrices ..thats ugly... must be some pattern!!

Oh. Now get what you meant by factoring .:P I tried that too and observed what you said :)

it seems transpose of itself!!

|dw:1336584742255:dw|
this is how far i got then..

This problem really is trollling me very badly :( I will solve this problem before going to sleeep

if you find let met know!!

Oo ...we can drop all ones!!

Oo ... i got zero!! LOL

@shivam_bhalla you still there??

Yes

No .. i know the shortcut too!!

actually i verified answer at wolframalpha!!

Ok. Just tell me the method and not the steps because that would literally troll you

take out all ones!!
sine they will have same values on row .. they will be zero ..

Sorry, didn't get you @experimentX

|dw:1336588777520:dw|

|dw:1336588827490:dw|

|dw:1336588893052:dw|

whenever matrix has same elements on row or column ... determinant is zero.

and by symmetry (b-c) and (a-c) are

;D

proportional ->eg R1 = kR2
so k = 1 is a specific case

Oo ... then i must have mistake!!

no its correct

if R1 = R2 then det(M) = 0

@eigenschmeigen , there is no R1=R2 in @experimentX 's proof

Oh my mistake ... i got the answer in fluke

@experimentX , I wish these flukes come for me too in my exam :P

LOL ... i haven't reviewed matrices in long time!!
best of luck if you are having exams right now!!

LOL. I have college entrance exam's right now :P

*exams

is this problem solved or still at large? i can have a proper go now i have a pen

The problem as of now is unsolved @eigenschmeigen

Okay ... i think i got this time!!
you can use same technique to reduce matrix!!

|dw:1336589808195:dw|

and then it's simplification!!

apply operation as above!!!
|dw:1336589933207:dw|

same as you had in your second place!! -- minus --- one

|dw:1336590227531:dw|

|dw:1336590458004:dw|
It can be proved!!

@experimentX , I will try and let you know in 5-10 minutes @experimentX

its B

i'm just checking my working

how do i write matrices on here?

@eigenschmeigen , your answer is right.
You can use drawing doard :)

ok gimme a sec

*board.

factor out (a-b)(b-c)

you can continue to factorise if you want, but this is sufficient to get the right option.

no problem, love matrices :)

what else does linear algebra cover?