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anonymous
 4 years ago
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??
anonymous
 4 years ago
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??

This Question is Closed

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0can you factor all of the elements

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus ,how exactly??

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[\Delta=\left\begin{array}\\1+a^2+a^4&1+ab+a^2b^2&1+ac+a^2c^2\\1+ab+a^2b^2&1+b^2+b^4&1+bc+b^2c^2\\1+ac+a^2c^2&1+bc+b^2c^2&1+c^2+c^4\end{array}\right\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0yeah maybe not, i really dont want to do it the normal way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Normal way>never :D It would take ages :P

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0but it would work in theory, but i would make about ten mistakes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am sure you play around with determinant properties like Row1 = Row 1 + Row 2+Row3 . I mean some tweaking of that sort

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0can we just do it the foolish way?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.0@apoorvk , I did that. But nothing can be taken common out :(

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0I'm going to try this problem

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0Have no fear, for inky is here!

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.1I got common stuff out through C1C2 and C2C3. but, i am still stuck with a trinomial in each ditch..

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0@shivam_bhalla I HAVE AN IDEA

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0aei+bfg+cdhcegbdiafh =a(eifh)+b(fgdj)+c(dgeg)

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0aei+bfg+cdhcegbdiafh aei+bf^2+cbfec^2ah^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@inkyvoyd , what is a,b,c,d,e,f,g?? Are you expanding the determinant??

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0I'm wikipediaing, and not making any progress.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL. @ "wikipediaing" . :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How to wolfram this ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@radar sir, any clue ??

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2factor out the commons!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX , how exactly ?? I mean row wise or column wise. And moreover, what to factor out ??

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2lol ... factoring out would lead to 3 more matrices ..thats ugly... must be some pattern!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh. Now get what you meant by factoring .:P I tried that too and observed what you said :)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2it seems transpose of itself!!

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1336584742255:dw this is how far i got then..

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2try 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
 4 years ago
Best ResponseYou've already chosen the best response.0This problem really is trollling me very badly :( I will solve this problem before going to sleeep

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2if you find let met know!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2Oo ...we can drop all ones!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2Oo ... i got zero!! LOL

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2@shivam_bhalla you still there??

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2looks 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
 4 years ago
Best ResponseYou've already chosen the best response.0LOL @experimentX , that is doing it the long way Atleast I know the answer know. :) Thanks for typing out the huge question on wolframalpha :)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2No .. i know the shortcut too!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2actually i verified answer at wolframalpha!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. Just tell me the method and not the steps because that would literally troll you

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2take out all ones!! sine they will have same values on row .. they will be zero ..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, didn't get you @experimentX

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1336588777520:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1336588827490:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1336588893052:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'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 (ab) must be a factor

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2whenever matrix has same elements on row or column ... determinant is zero.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and by symmetry (bc) and (ac) are

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.0proportional >eg R1 = kR2 so k = 1 is a specific case

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2Oo ... then i must have mistake!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if R1 = R2 then det(M) = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@eigenschmeigen , there is no R1=R2 in @experimentX 's proof

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2Oh my mistake ... i got the answer in fluke

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX , I wish these flukes come for me too in my exam :P

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2LOL ... i haven't reviewed matrices in long time!! best of luck if you are having exams right now!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL. I have college entrance exam's right now :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is this problem solved or still at large? i can have a proper go now i have a pen

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The problem as of now is unsolved @eigenschmeigen

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2Okay ... i think i got this time!! you can use same technique to reduce matrix!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1336589808195:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2and then it's simplification!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2apply operation as above!!! dw:1336589933207:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2same as you had in your second place!!  minus  one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1336590227531:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1336590458004:dw It can be proved!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX , I will try and let you know in 510 minutes @experimentX

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2okay ... 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
 4 years ago
Best ResponseYou've already chosen the best response.0i'm just checking my working

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how do i write matrices on here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@eigenschmeigen , your answer is right. You can use drawing doard :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.0letting 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}a^4 + a^2  ab  a^2b^2 & a^2b^2 + ab b^2 b^4 & a^2c^2 + ac  bc  b^2c^2 \\ a^2b^2 + ab ac a^2c^2 & b^4 + b^2  bc  b^2c^2 & b^2c^2 + bc c^2  c^4 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0factor out (ab)(bc)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(ab)(bc)\left[\begin{matrix}a^3 + a +ba^2 & b^3 + b + ab^2 & c(ac+bc+1) \\ a(ac+ab+1) & b^3 + b +cb^2 & c^3 + c + bc^2 \\ 1 +ac +a^2c^2 & 1 +bc+b^2c^2 & 1 + c^2 + c^4\end{matrix}\right]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0letting a = b we now have C1 = C2 so we can see that (ab) is again a factor. this narrows it down to B as we will have (ab)^2 in our answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can continue to factorise if you want, but this is sufficient to get the right option.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its 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
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you @eigenschmeigen and @experimentX for both of your solutions :D Thanks for taking your time :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no problem, love matrices :)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2looks 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
 4 years ago
Best ResponseYou've already chosen the best response.0what else does linear algebra cover?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2well ... 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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