anonymous
  • anonymous
Ok. A determinants question(Link below). What I have done? Well, I tried Row1 = Row 1+Row 2+row 3 but nothing great. Can you guys guide me to get the answer??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
http://mypat.in/PaperImages/3/425/Q_4_004.gif
UnkleRhaukus
  • UnkleRhaukus
can you factor all of the elements
anonymous
  • anonymous
@UnkleRhaukus ,how exactly??

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UnkleRhaukus
  • UnkleRhaukus
\[\Delta=\left|\begin{array}\\1+a^2+a^4&1+ab+a^2b^2&1+ac+a^2c^2\\1+ab+a^2b^2&1+b^2+b^4&1+bc+b^2c^2\\1+ac+a^2c^2&1+bc+b^2c^2&1+c^2+c^4\end{array}\right|\]
UnkleRhaukus
  • UnkleRhaukus
yeah maybe not, i really dont want to do it the normal way
anonymous
  • anonymous
Normal way-->never :D It would take ages :P
UnkleRhaukus
  • UnkleRhaukus
but it would work in theory, but i would make about ten mistakes
anonymous
  • anonymous
I am sure you play around with determinant properties like Row1 = Row 1 + Row 2+Row3 . I mean some tweaking of that sort
anonymous
  • anonymous
@apoorvk
inkyvoyd
  • inkyvoyd
can we just do it the foolish way?
apoorvk
  • apoorvk
Hmm, tried C1 - C2?
anonymous
  • anonymous
@inkyvoyd Sorry, please, In my exam I have only 3 min(max) to solve this question @apoorvk ,let me give it a try and get back to you
anonymous
  • anonymous
@apoorvk , I did that. But nothing can be taken common out :(
inkyvoyd
  • inkyvoyd
I'm going to try this problem
inkyvoyd
  • inkyvoyd
Have no fear, for inky is here!
apoorvk
  • apoorvk
I got common stuff out through C1-C2 and C2-C3. but, i am still stuck with a trinomial in each ditch..
inkyvoyd
  • inkyvoyd
@shivam_bhalla I HAVE AN IDEA
inkyvoyd
  • inkyvoyd
aei+bfg+cdh-ceg-bdi-afh =a(ei-fh)+b(fg-dj)+c(dg-eg)
inkyvoyd
  • inkyvoyd
b=d f=h
inkyvoyd
  • inkyvoyd
c=g
inkyvoyd
  • inkyvoyd
aei+bfg+cdh-ceg-bdi-afh aei+bf^2+cbf-ec^2-ah^2
anonymous
  • anonymous
@inkyvoyd , what is a,b,c,d,e,f,g?? Are you expanding the determinant??
inkyvoyd
  • inkyvoyd
I'm wikipediaing, and not making any progress.
anonymous
  • anonymous
LOL. @ "wikipediaing" . :)
apoorvk
  • apoorvk
wolfram!!!
anonymous
  • anonymous
How to wolfram this ??
anonymous
  • anonymous
@radar sir, any clue ??
experimentX
  • experimentX
factor out the commons!!!
anonymous
  • anonymous
@experimentX , how exactly ?? I mean row wise or column wise. And moreover, what to factor out ??
experimentX
  • experimentX
lol ... factoring out would lead to 3 more matrices ..thats ugly... must be some pattern!!
anonymous
  • anonymous
Oh. Now get what you meant by factoring .:P I tried that too and observed what you said :)
experimentX
  • experimentX
it seems transpose of itself!!
apoorvk
  • apoorvk
|dw:1336584742255:dw| this is how far i got then..
experimentX
  • experimentX
try making these types to a^4 + a^2 + 1 => a^6 - 1 (ab)^2 + (ab) + 1 => (ab)^3 - 1 ... not sure if it can help ... too tired!!!
anonymous
  • anonymous
This problem really is trollling me very badly :( I will solve this problem before going to sleeep
experimentX
  • experimentX
if you find let met know!!
experimentX
  • experimentX
Oo ...we can drop all ones!!
experimentX
  • experimentX
Oo ... i got zero!! LOL
experimentX
  • experimentX
@shivam_bhalla you still there??
anonymous
  • anonymous
Yes
experimentX
  • experimentX
looks like it got it correct http://www.wolframalpha.com/input/?i=det+{{1%2Ba^2%2Ba^4%2C+1+%2B+%28ab%29+%2B+%28ab%29^2%2C+1+%2B+%28ac%29+%2B+%28ac%29^2}%2C++%09{1+%2B+%28ab%29+%2B+%28ab%29^2%2C+1%2Bb^2%2Bb^4%2C+1+%2B+bc+%2B+%28bc%29^2}%2C++%09{1+%2B+%28ac%29+%2B+%28ac%29^2%2C+1+%2B+bc+%2B+%28bc%29^2%2C+1+%2B+c^2+%2B+c^4}}
anonymous
  • anonymous
LOL @experimentX , that is doing it the long way Atleast I know the answer know. :) Thanks for typing out the huge question on wolframalpha :)
experimentX
  • experimentX
No .. i know the shortcut too!!
experimentX
  • experimentX
actually i verified answer at wolframalpha!!
anonymous
  • anonymous
Ok. Just tell me the method and not the steps because that would literally troll you
anonymous
  • anonymous
@experimentX
experimentX
  • experimentX
take out all ones!! sine they will have same values on row .. they will be zero ..
anonymous
  • anonymous
Sorry, didn't get you @experimentX
experimentX
  • experimentX
|dw:1336588777520:dw|
experimentX
  • experimentX
|dw:1336588827490:dw|
anonymous
  • anonymous
|dw:1336588893052:dw|
anonymous
  • anonymous
i'm not sure how far everyone has got so far but first thoughts: setting a = b we can see that the determinant is zero as R1 = R2 hence (a-b) must be a factor
experimentX
  • experimentX
whenever matrix has same elements on row or column ... determinant is zero.
anonymous
  • anonymous
and by symmetry (b-c) and (a-c) are
anonymous
  • anonymous
@experimentX , I never studied that. I only studied that if two rows or 2 colums are proportional to each other, then the determinant of the matrix is zero @eigenschmeigen , Very logical approach which eliminates option (a) for us :)
anonymous
  • anonymous
;D
anonymous
  • anonymous
proportional ->eg R1 = kR2 so k = 1 is a specific case
experimentX
  • experimentX
Oo ... then i must have mistake!!
anonymous
  • anonymous
no its correct
anonymous
  • anonymous
if R1 = R2 then det(M) = 0
anonymous
  • anonymous
@eigenschmeigen , there is no R1=R2 in @experimentX 's proof
experimentX
  • experimentX
Oh my mistake ... i got the answer in fluke
anonymous
  • anonymous
@experimentX , I wish these flukes come for me too in my exam :P
experimentX
  • experimentX
LOL ... i haven't reviewed matrices in long time!! best of luck if you are having exams right now!!
anonymous
  • anonymous
LOL. I have college entrance exam's right now :P
anonymous
  • anonymous
*exams
anonymous
  • anonymous
is this problem solved or still at large? i can have a proper go now i have a pen
anonymous
  • anonymous
The problem as of now is unsolved @eigenschmeigen
experimentX
  • experimentX
Okay ... i think i got this time!! you can use same technique to reduce matrix!!
experimentX
  • experimentX
|dw:1336589808195:dw|
experimentX
  • experimentX
and then it's simplification!!
experimentX
  • experimentX
apply operation as above!!! |dw:1336589933207:dw|
experimentX
  • experimentX
same as you had in your second place!! -- minus --- one
anonymous
  • anonymous
|dw:1336590227531:dw|
experimentX
  • experimentX
1 Attachment
experimentX
  • experimentX
|dw:1336590458004:dw| It can be proved!!
anonymous
  • anonymous
@experimentX , I will try and let you know in 5-10 minutes @experimentX
experimentX
  • experimentX
okay ... it's just simplification after that. factoring out common factor ... then again you will have 1's and then use same technique ... in the end you will get two linearly dependent columns!! then it will be zero!!
anonymous
  • anonymous
its B
anonymous
  • anonymous
i'm just checking my working
anonymous
  • anonymous
how do i write matrices on here?
anonymous
  • anonymous
@eigenschmeigen , your answer is right. You can use drawing doard :)
anonymous
  • anonymous
ok gimme a sec
anonymous
  • anonymous
*board.
anonymous
  • anonymous
i like the equation writer, i'm quicker that way: \[\left[\begin{matrix}1+a^2 + a^4 & 1+ ab+a^2b^2 & 1 +ac +a^2c^2 \\ 1 + ab +a^2b^2 & 1+b^2 + b^4 & 1 +ac +a^2c^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]
anonymous
  • anonymous
letting a = b we get R1 = R2 letting b = c we get R2 = R3 so the operations to do are R1- R2 = R1' and R2 - R3 = R2'
anonymous
  • anonymous
\[\left[\begin{matrix}a^4 + a^2 - ab - a^2b^2 & a^2b^2 + ab -b^2 -b^4 & a^2c^2 + ac - bc - b^2c^2 \\ a^2b^2 + ab -ac -a^2c^2 & b^4 + b^2 - bc - b^2c^2 & b^2c^2 + bc -c^2 - c^4 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]
anonymous
  • anonymous
factor out (a-b)(b-c)
anonymous
  • anonymous
\[(a-b)(b-c)\left[\begin{matrix}a^3 + a +ba^2 & b^3 + b + ab^2 & c(ac+bc+1) \\ a(ac+ab+1) & b^3 + b +cb^2 & c^3 + c + bc^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]
anonymous
  • anonymous
letting a = b we now have C1 = C2 so we can see that (a-b) is again a factor. this narrows it down to B as we will have (a-b)^2 in our answer
anonymous
  • anonymous
you can continue to factorise if you want, but this is sufficient to get the right option.
anonymous
  • anonymous
its not said in many textbooks but i find using factor theorem is really useful for factorising determinants, as it tells us an operation that will result in something we can factor
anonymous
  • anonymous
Thank you @eigenschmeigen and @experimentX for both of your solutions :D Thanks for taking your time :)
anonymous
  • anonymous
no problem, love matrices :)
experimentX
  • experimentX
looks like i didn't get that right ... i'm embarrassed. looks like 'linear algebra' is the first thing i should do after i'm done with calc
anonymous
  • anonymous
what else does linear algebra cover?
experimentX
  • experimentX
well ... all forgot!! let's see ... i have a book here. Linear algebra - Serge Lang Vectors, Vector spaces, Matirces, Linear mappings, Linear mappings and matrices, .... Groups, Rings, Convex Sets ... Appendix I thought \( \Delta\) on wolfram was the answer!!

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